Basic Math Challenge - August 2018

[fresh_42]

I'm not saying you should write problems for me, just noting that age might not be the right qualifier to use. I did expect something a little simpler when I saw "Basic Math Challenge"... :-)

I basically agree on what you wrote, and the age criterion referred to the "usual development" given by school education. But you are right, it is not generally a good one. I would even challenge it from the other side: Just because a problem is given in a group theoretic language, doesn't automatically make it hard, since high schoolers might well be capable to solve it once they've cleared terminological ground. And yes, it is true that some of the intermediate problems are rated by their language rather than their difficulty. Funny enough, some are still unsolved, although an hour of Wikipedia reading would probably be all what is needed to overcome this hurdle.

However, the main concern remains unsolved: how to prevent those who are well trained from immediately solving them? And if not by age, how should we define "basic"? It differs a lot depending on whom you ask.

 

[Freixas]

Spoiler: Comment on Solution to 9

Since her angular velocity is higher than that of the witch she must reach such a point by the intermediate value theorem.

What a great solution! I found I could visualize the answer more easily this way:

The princess can swim a circle with a circumference of .48π in, say, 1 unit of time. The point opposite her, on shore, moves at 2π in the same amount of time, or at 4.1666.. times the princess's speed. The witch's job is now to avoid this moving location, but at best, she can only move at 4 times the princess's speed.

To finish off, you could assume the princess swims in just one direction, clockwise. The witch's maximal distance from the dreaded "opposite" point is just shy of 2π (as seen by the point, which only looks in the clockwise direction). If the witch runs clockwise, she falls behind, which means the forward distance to the witch (as seen by the point) is reduced. If the witch runs counter-clockwise, the distance is clearly reduced. The argument holds at any distance from the point.

 

[Freixas]

However, the main concern remains unsolved: how to prevent those who are well trained from immediately solving them? And if not by age, how should we define "basic"?

Ignoring age, what is "basic"? You can teach a 10-year old or a 40-year old (who skipped school) the same "basic" math concepts. Maybe you approach it from the point of view of math knowledge required (I'm thinking of things like arithmetic, geometry, algebra, calculus, etc.).

To keep it simple, you might just tag the truly beginner problems by describing who you think the problem is targeted to. If you include the word "beginner" somewhere in there, you might you might solve the first problem, too.

I do think the princess problem could be solved using beginning geometry, but it's still not an easy problem. So there's that, too.

 

[fresh_42]

I do think the princess problem could be solved using beginning geometry

All you need formally are the equations circumference equals two $\pi$ times radius and the definition of angular velocity, both very basic concepts.

 

[lpetrich]

In my solution to Question 2, I made a mistake in my statement of the generalized binomial theorem. Here is the correct statement of it:
$$ (1+x)^a = 1 + a x + \frac{a(a-1)}{2!} x^2 + \frac{a(a-1)(a-2)}{3!} x^3 + \dots = \sum_{k=0}^\infty \frac{\Gamma(a+1)}{\Gamma(a+1-k) k!} $$
using the Euler gamma function.

 

[Freixas]

All you need formally are the equations circumference equals two $\pi$ times radius and the definition of angular velocity, both very basic concepts.

The statement of mine that you quoted reflects that we're in agreement. I go along with the claim that, for a member of PhysicsForums, these can assumed to be basic concepts. Not that the assumption is always correct; just that you have to set some baseline.

I said it's not an easy problem based on this line in the original problem statement:

Why doesn't she have a chance to escape?

In fact, she does have a chance, as was nicely proved by Danny Sleator. I figured if you (or whoever wrote this), thought that the princess had no chance, it must, by definition, not be an easy problem. You might want to edit the problem statement, unless it was meant as a deliberate red herring.

 

[StoneTemplePython]

Most of the problems listed here look like gobbledygook to me. The princess one is about the only one I might have tackled (OK, I'm lazy). When I was in high school I was considered one of the top students in my math class. In college, I took advanced math classes and got A's...

I did expect something a little simpler when I saw "Basic Math Challenge"... :-)

what about the gas canisters problem in our last basic challenge? It was problem 2:

https://www.physicsforums.com/threads/basic-math-challenge-july-2018.950689/

= = = =
I think it garnered a lot of interest as it is understandable to most everyone. There were a lot of attempts at it. @mfb solved it in perhaps the most constructive way (post number 48) - - - however his discussions of global minimum, etc. could be off-putting for some I suppose.

A simpler approach is to solve it inductively in 3 lines -- though I'll show it in 5 lines via iteration. sketch:

Spoiler

1.) total fuel supply exactly matches the total fuel required by track length so either we have the special case where all intervals exactly have the amount of gas required to complete the interval (in which case the problem is done) or there must be at least one interval that has more fuel than distance traveled. For Basic thread people can reason this out -- say it's obvious with a couple sentences explaining-- I thought this insight was, essentially, had in post 26. A tighter approach would be to show this with some inequalities but I wouldn't insist on it necessarily for Basic.

2.) For the $n= 2$ canisters case, start at the one that has a surplus -- i.e. the canister that has more fuel than the distance from it to the next canister. (Again if everything is zero balance, we can choose any starting spot and we're done.). We now know a viable starting position exists for all $n=2$ canister cases.

3.) For the 3 canisters case (and again if everything is zero balance we're done, so assume it's not) use bunching: We know there is at least one canister with a surplus and so we can bunch/fuse it with the canister sequentially after it. Because there's a fuel surplus in between the two there's no danger in running out of gas in that interval -- so we can just consider total distance and total fuel change in the 'bunched' setup. But now we have a 2 canister case where we've already showed that a viable starting point exists. Hence we know a viable starting position exists for all $n=3$ canister cases.

4.) For the 4 canisters case (and again if everything is zero balance we're done, so assume it's not) use bunching: Consider the canister with the surplus and bunch/fuse it with the canister sequentially after it. We've now reduced the problem to the 3 canister case which we've already solved in the affirmative... so we now know that a viable starting position must exist for all $n=4$ canister cases.

5.) and so on...

- - -
so yes there must always be some way of driving over the track without running out of fuel in the $n$ canister case.

edit:
I'm not sure if it's helpful, but you can also view this in terms of an $n=1$ gas canisters on the track must be trivially true / by definition you pick that can start at the single gas canister and have just enough gas to make the lap. You could then view, if you like, the $n=2$ case as a case where you select the canister with surplus and via bunching have now reduced the problem to the $n=1$ case. Doing the recurrence at the $n=2$ case something helps with people's intuition, other times I think they want to directly solve the $n=2$ case and set it as the base case.

- - - -

Ignoring age, what is "basic"? You can teach a 10-year old or a 40-year old (who skipped school) the same "basic" math concepts...

I do think the princess problem could be solved using beginning geometry, but it's still not an easy problem. So there's that, too.

The only technique needed for the above gas canisters problem is familiarity with induction (or iteration), and to play around with the problem for a bit. In this sense the problem is Basic as induction is perhaps the most basic proof technique -- but the emphasis is on Basic Math Challenge -- i.e. people need to play around with the problem for a while because it is a challenge.

 

[fresh_42]

unless it was meant as a deliberate red herring.

It was. One lesson to learn is not to have prejudices anywhere: Rely on your argument, not on what's written, it might be wrong.

 

[archaic]

What's the difference between $\mathbb{N}$ and $\mathbb{N}_0$ ? Wikipedia says it's the same. (reference to the first question)

 

[fresh_42]

What's the difference between $\mathbb{N}$ and $\mathbb{N}_0$ ? Wikipedia says it's the same. (reference to the first question)

I used the distinction $\mathbb{N}=\{\,1,2,3,\ldots \,\}\, , \,\mathbb{N}_0=\{\,0,1,2,3,\ldots \,\}$.
It's unnecessary to discuss whether zero is a natural number or not. Both have pros and cons.

 

[lpetrich]

I used the distinction $\mathbb{N}=\{\,1,2,3,\ldots \,\}\, , \,\mathbb{N}_0=\{\,0,1,2,3,\ldots \,\}$.
It's unnecessary to discuss whether zero is a natural number or not. Both have pros and cons.

One could use $\mathbb{N}_1$ instead of $\mathbb{N}$ to indicate its starting number, as with $\mathbb{N}_0$.

 

[Chris Miller]

Most involve terminologies that prevent my understanding the problem. Question 9 was fun though. Something to think about. I'd advise the princess to swim in the smallest possible circle in the center of her pond, forcing the witch to sprint the pond's circumference in order to maintain minimal distance. Even if she's a Kenyan marathoner, she should drop dead in a few hours. Failing this, I'd advise she swim away from the witch in the direction that is perpendicular to the tangent of the point the witch is on, which will be a curve if the witch is moving along the circumference. In other words, to maintain maximum distance from the witch at all times.

 

[mfb]

One could use $\mathbb{N}_1$ instead of $\mathbb{N}$ to indicate its starting number, as with $\mathbb{N}_0$.

I have seen $\mathbb{N}^+$ for natural numbers starting at 1.

 

[member 587159]

My solution for (8):

Spoiler

(a) Since $F$ is a distribution function, it must satisfy $F \to 1$ whenever $x \to \infty$. Hence, $a=1$.

Since $X$ denotes waiting time, and it isn't physically possible to wait a negative time, we have $\mathbb{P}\{X < 0\} = \mathbb{P}( \emptyset) = 0$. It follows that $F(0) = \mathbb{P}\{X \leq 0\} = \mathbb{P}\{X < 0\} + \mathbb{P}\{X = 0\} = 1/2$, so $a -b = 1-b = 1/2$ and $b = 1/2$.

On the other hand, $1 - F(1) = 1 - \mathbb{P}\{X \leq 1\} = \mathbb{P}\{X > 1\} = 1/4$. It follows that $3/4 = F(1) = 1-1/2e^{- \lambda}$. This can be solved for $\lambda$ with the help of logarithms.

(b) The distribution can be described by a density function, because the distribution function is differentiable everywhere except possibly not at $0$. Hence, we can define the density function $f$ by

$f(x) = \begin{cases} 0 \quad x = 0 \\ b \lambda e^{- \lambda x} \quad x \neq 0\end{cases}$

Here, I used a basic result in probability theory: if a distribution function is differentiable everywhere, except at an (at most) countable set of points, the distribution is absolutely continuous, and the density is given by the derivative on the points where the function is differentiable and 0 elsewhere.

 

[fresh_42]

My solution for (8):
(a) Since $F$ is a distribution function, it must satisfy $F \to 1$ whenever $x \to \infty$. Hence, $a=1$.

Since $X$ denotes waiting time, and it isn't physically possible to wait a negative time, we have $\mathbb{P}\{X < 0\} = \mathbb{P}( \emptyset) = 0$. It follows that $F(0) = \mathbb{P}\{X \leq 0\} = \mathbb{P}\{X < 0\} + \mathbb{P}\{X = 0\} = 1/2$, so $a -b = 1-b = 1/2$ and $b = 1/2$.

On the other hand, $1 - F(1) = 1 - \mathbb{P}\{X \leq 1\} = \mathbb{P}\{X > 1\} = 1/4$. It follows that $3/4 = F(1) = 1-1/2e^{- \lambda}$. This can be solved for $\lambda$ with the help of logarithms.

So?

(b) The distribution can be described by a density function, because the distribution function is differentiable everywhere except possibly not at $0$. Hence, we can define the density function $f$ by

$f(x) = \begin{cases} 0 \quad x = 0 \\ b \lambda e^{- \lambda x} \quad x \neq 0\end{cases}$

Here, I used a basic result in probability theory: if a distribution function is differentiable everywhere, except at an (at most) countable set of points, the distribution is absolutely continuous, and the density is given by the derivative on the points where the function is differentiable and 0 elsewhere.

Could you give a quotation of this "basic result"?
But anyway, this is the "Basic" challenge, so we do not talk about measure theory and integrability means Riemann, not Lebesgue. However, as you pointed out the critical point $x=0$ it is a correct answer, however, in basic math terms there is no density function because of the discontinuity at $x=0\,.$

Summary:

Basic: Because the distribution function isn't continuous (at $x=0$) and thus not differentiable, it cannot result from a density function.
(acceptable answer in this case)

Advanced mathematics: The discontinuity at $x=0$ isn't a satisfactory hurdle for not to speak of a density function, which becomes obvious if we draw the graph. To overcome such obstacles is the main reason why probability theory nowadays is based on measure theory and Lebesgue integration, rather than combinatorics and Riemann integrability. So in case you encounter $\sigma-$algebras somewhere, this is the reason: make sets measurable (=volume) in a meaningful way, although they might contain a few problematic points. It means: Make the best out of the misery, as long as we still can get a well-defined theory. It's like having a ball minus a point or a two-dimensional slice, in which case the ball has still the volume $\frac{4}{3}\pi r^3\,.$ That's the basic idea behind measure theory and the related $\sigma-$algebras, so don't be scared.

 

[member 587159]

If I made no mistakes $\lambda = \ln(2)$

So?

Could you give a quotation of this "basic result"?

I thought it was basic because it was mentioned in my probability theory book from my university, but it does mention that the proof is not trivial. So maybe basic theorem, but certainly not a basic proof. I also don't seem to find a good source to back up. When university starts again, I will ask my probability prof for a reference.

 

[fresh_42]

If I made no mistakes $\lambda = \ln(2)$

Yes.
$$
F(x) =
\begin{cases}
0 & \text{if } x < 0 \\
1-\dfrac{1}{2}e^{-\log 2\,x} & \text{if } x \geq 0
\end{cases}
$$

I thought it was basic because it was mentioned in my probability theory book from my university, but it does mention that the proof is not trivial. So maybe basic theorem, but certainly not a basic proof. I also don't seem to find a good source to back up. When university starts again, I will ask my probability prof for a reference.

If we take the Wikipedia definition then we have integrability (Lebesgue) and we are led into measure theory. So all depends on whether continuity is required or not. As it commonly is, the answer should be no.

 

[fresh_42]

Solution for problem #4:

Since @lpetrich's solution doesn't contain the calculations for the integrals, used confusingly different letters, and I have made a typo by defining the paths, I now add a complete solution for problem #4 which also has a bit more explicit reason for the question about a possible potential of the two vector fields:

In the first step we parameterize the two curves:
\begin{align*}
c_1\, &: \,\left[ -\frac{\pi}{2},\frac{\pi}{2} \right] \longrightarrow \gamma_1 \text{ with } c_1(t)=(\cos t,\sin t) \text{ for } \gamma_1 \text{ and } \\
c_{2,1}\, &: \,[0,1] \longrightarrow\gamma_2 \text{ with } c_{2,1}(t)=(t,t-1) \text{ and }\\
c_{2,2}\, &: \,[0,1] \longrightarrow\gamma_2 \text{ with } c_{2,2}(t)=(1-t,t) \text{ for } \gamma_2
\end{align*}
and get $\dot{c}_{1}(t)=(-\sin t, \cos t)\; , \;\dot{c}_{2,1}(t)=(1,1)\; , \;\dot{c}_{2,2}(t)=(-1,1)\,.$ Thus
\begin{align*}
\int_{\gamma_1} v\,ds &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} v(c_1(t)) \cdot \dot{c}_1(t) \,dt \\
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \begin{bmatrix}
\sin t \\ \cos t -\sin t
\end{bmatrix} \cdot \begin{bmatrix}
-\sin t \\ \cos t
\end{bmatrix}\, dt\\
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} -\sin^2 t + \cos^2 t - \sin t \cos t \,dt\\
&=\left[ -\frac{1}{2}\cos^2 t + 2\sin t \cos t \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\\
&=0
\end{align*}
\begin{align*}
\int_{\gamma_2} v\,ds &= \int_0^1 v(c_{2,1}(t)) \cdot \dot{c}_{2,1}(t) \,dt + \int_0^1 v(c_{2,2}(t)) \cdot \dot{c}_{2,2}(t) \,dt \\
&= \int_0^1 \begin{bmatrix}
t-1\\1 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 1 \end{bmatrix} + \begin{bmatrix} t \\1-2t \end{bmatrix} \cdot \begin{bmatrix} -1\\1 \end{bmatrix} \,dt \\
&= \int_0^1 1-2t \,dt\\
&= \left[ t-t^2 \right]_0^1 \\
&= 0
\end{align*}
\begin{align*}
\int_{\gamma_1} w\,ds &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} w(c_1(t)) \cdot \dot{c}_1(t) \,dt \\
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \begin{bmatrix}
\sin t - \cos t \\ -\sin t
\end{bmatrix} \cdot \begin{bmatrix}
-\sin t \\ \cos t
\end{bmatrix}\, dt\\
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} -\sin^2 t \,dt\\
&=\left[ -\frac{1}{2} (t - \sin t \cos t ) \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\\
&=-\frac{\pi}{2}
\end{align*}
\begin{align*}
\int_{\gamma_2} w\,ds &= \int_0^1 w(c_{2,1}(t)) \cdot \dot{c}_{2,1}(t) \,dt + \int_0^1 w(c_{2,2}(t)) \cdot \dot{c}_{2,2}(t) \,dt \\
&= \int_0^1 \begin{bmatrix} -1 \\ 1-t
\end{bmatrix} \cdot \begin{bmatrix} 1 \\ 1 \end{bmatrix} + \begin{bmatrix} 2t-1 \\ -t \end{bmatrix} \cdot \begin{bmatrix} -1\\1 \end{bmatrix} \,dt \\
&= \int_0^1 1-4t \,dt\\
&= \left[ t-2t^2 \right]_0^1 \\
&= -1
\end{align*}
So the vector field $v$ is apparently path independent whereas $w$ is not. We check this by the calculation of their curl.
$$\operatorname{rot}\vec{F}(x,y) = \operatorname{curl}\vec{F}(x,y) =
\dfrac{\partial \vec{F}_y}{\partial x} - \dfrac{\partial \vec{F}_x}{\partial y} =
\begin{cases}
0 & \text{if } \vec{F} = v \\
-1 & \text{if } \vec{F} = w
\end{cases}$$
So $v$ has a potential and thus is path independent, and $w$ has none.

 

[member 587159]

So?

Could you give a quotation of this "basic result"?
But anyway, this is the "Basic" challenge, so we do not talk about measure theory and integrability means Riemann, not Lebesgue. However, as you pointed out the critical point $x=0$ it is a correct answer, however, in basic math terms there is no density function because of the discontinuity at $x=0\,.$

Summary:

Basic: Because the distribution function isn't continuous (at $x=0$) and thus not differentiable, it cannot result from a density function.
(acceptable answer in this case)

Advanced mathematics: The discontinuity at $x=0$ isn't a satisfactory hurdle for not to speak of a density function, which becomes obvious if we draw the graph. To overcome such obstacles is the main reason why probability theory nowadays is based on measure theory and Lebesgue integration, rather than combinatorics and Riemann integrability. So in case you encounter $\sigma-$algebras somewhere, this is the reason: make sets measurable (=volume) in a meaningful way, although they might contain a few problematic points. It means: Make the best out of the misery, as long as we still can get a well-defined theory. It's like having a ball minus a point or a two-dimensional slice, in which case the ball has still the volume $\frac{4}{3}\pi r^3\,.$ That's the basic idea behind measure theory and the related $\sigma-$algebras, so don't be scared.

Revisiting this thread, I feel like some things should be cleared out if future readers read these posts.

From a measure theoretic viewpoint, it is necessary that a distribution function $F$ is continuous if we want an absolutely continuous distribution for a random variable $X$. This, because $F$ is continuous in $x$ if and only if $P(X = x) = 0$ and if our distribution is continuous, we can write
$\mu(A) = \int_A f d \lambda$ where $\mu]-\infty, x]= F(x)$. In particular $P(X= x)$ must be $0$ for all $x$.

So in any case, there is no density. Also not if we consider measure theoretic stuff.

The result I quoted is the following:

Let $\mu: \mathcal{R} \to [0,1]$ be a probability measure on the Borel sets of the real numbers such that the distribution function $F(x) = \mu]-\infty,x]$ is continuous and differentiable everywhere except in an at most countable set. Then $\mu$ is absolutely continuous, i.e. the distribution function $F$ has a density. It is in most measure theory books that treat differentiation, but maybe not in the probability theoretic form like I wrote it down here.

Because of the discontinuity in 0, it was not applicable.

 

[MAGNIBORO]

Solution for problem #1:
Let $$\sum 2^{n}x_{2^{n}}=A \; \; \; \, \, \: \: \: \: ,\sum x_{n}=B\; \; \; \, \, \: \: \: \: , \sum 2^{n-1}x_{2^{n}}=C $$
If we assume a $a_{n}$ such that $A$ diverge and $B$ converge
we derive to a contradiccion; so:

because $a_{n}$ is monotone decreasent, is easy to see that $$A>B>C$$ and because B converge then C must converges as well.

but $$A-C=C$$ $$A=2C$$

this is a contradiccion , because some constant multiplied with a convergent series cannot be a divergent series.

then cannot exist a $a_{n}$ that make $A$ diverge and $B$ converge.
finally, the convergance of $B$ implies the convergance of $A$ and viceversa

 

[fresh_42]

because $a_{n}$ is monotone decreasent, is easy to see that
$$A>B>C$$

This is the crucial point and "easy to see" shouldn't occur in the Basic Challenge.

 

[Periwinkle]

1. Given a non-negative, monotone decreasing sequence $(a_n)_{n \in \mathbb{N}}\subseteq \mathbb{R}\,.$ Prove that $\sum_{n \in \mathbb{N}}a_n$ converges if and only if $\sum_{n \in \mathbb{N}_0}2^na_{2^n}$ converges.

$$S = \sum_{n \in \mathbb{N}}a_n \\ W = \sum_{n \in \mathbb{N}_0}2^na_{2^n} $$

If $~W$ converges then $S$ also converges​

$$ a_1 \leq a_1 \\ a_2 + a_3 \leq 2a_2 \\ a_4 + a_5+a_6 + a_7 \leq 4a_4 \\ \dots \\a_{2^n} + a_{2^n+1}+ \dots + a_{2^{n+1}-1} \leq 2^n a_{2^n} \\ \dots$$

If $~S$ converges then $W$ also converges​

$$ 2a_1 \gt a_1 \\ 2a_2 \geq 2 a_2 \\ 2(a_3+a_4) \geq 4 a_4 \\ 2(a_5+a_6+a_7+a_8) \geq 8a_8 \\ \dots \\ 2(a_{2^n+1}+a_{2^n+2}+ \dots +a_{2^{n+1}}) \geq 2^{n+1}a_{2^{n+1}} \\ \dots$$

 

[fresh_42]

$$S = \sum_{n \in \mathbb{N}}a_n \\ W = \sum_{n \in \mathbb{N}_0}2^na_{2^n} $$

If $~W$ converges then $S$ also converges​

$$ a_1 \leq a_1 \\ a_2 + a_3 \leq 2a_2 \\ a_4 + a_5+a_6 + a_7 \leq 4a_4 \\ \dots \\a_{2^n} + a_{2^n+1}+ \dots + a_{2^{n+1}-1} \leq 2^n a_{2^n} \\ \dots$$

If $~S$ converges then $W$ also converges​

$$ 2a_1 \gt a_1 \\ 2a_2 \geq 2 a_2 \\ 2(a_3+a_4) \geq 4 a_4 \\ 2(a_5+a_6+a_7+a_8) \geq 8a_8 \\ \dots \\ 2(a_{2^n+1}+a_{2^n+2}+ \dots +a_{2^{n+1}}) \geq 2^{n+1}a_{2^{n+1}} \\ \dots$$

This is correct, as it is in here (I know you didn't see it earlier):
https://www.physicsforums.com/threads/basic-math-challenge-september-2018.954490/#post-6051193And you deserve the same comment, too :
Your proof is correct, although I think that working with partial sums instead of infinite many dots would have been a bit more professional. Here is what I mean (same argument, just written a bit differently):

Cauchy's Condensation criterion.
Let's assume $\sum_{n \in \mathbb{N}}a_n$ converges. We set $S_{n}=\sum_{k=1}^{n}a_k$ and calculate
\begin{align*}
S_{2^n} &\geq a_1+a_2+2a_4+4a_8+ \ldots +2^{n-1}a_{2^n}\\
&\geq \frac{1}{2}\left(a_1+2a_2+4a_4+8a_8+\ldots +2^na_{2^n}\right)\\
&=\frac{1}{2}\sum_{k=0}^{2^n}2^ka_k
\end{align*}
Since $\sum_{k=1}^{\infty} a_k$ converges, so does the series $(S_n)_{n \in \mathbb{N}}$ of partial sums and thus twice the subsequence $2\cdot(S_{2^n})_{n \in \mathbb{N}}$. But this is the boundary from above for the non-negative sums $\sum_{k=1}^{n}2^ka_k$, i.e. $\sum_{k=1}^{\infty}2^ka_k$ converges.

Let now $n<2^{m+1}-1\,.$ Then
\begin{align*}
\sum_{k=1}^{n}a_k&\leq \sum_{k=1}^{2^{m+1}-1}a_k\\
&\leq a_1 + (a_2+a_2)+(a_4+a_4+a_4+a_4)+(a_8+\ldots)+(a_{2^m}+\ldots)\\
&=\sum_{k=0}^{m}2^ka_k
\end{align*}
If $\sum_{k=0}^{\infty}2^ka_k$ converges, then $\sum_{k=0}^{\infty}a_k$ is bounded and converges, too.