Fun Puzzles: Test Your Math Skills!

[sbhatnagar]

Fun Problems! Evaluate the following:

1. \( \displaystyle \sum_{n=1}^{\infty}\frac{\left( 1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots +\dfrac{1}{(n-1)!} \right)}{2^{n-1}}\)

2. \( \displaystyle \sum_{n=1}^{\infty}\frac{1+2+2^2+\cdots+2^{n-1}}{n!}\)

3. \( \displaystyle \sum_{n=1}^{\infty}\frac{(1+3^n)\ln^n(3)}{n!}\)

4. \( \displaystyle \sum_{n=1}^{\infty}\dfrac{\displaystyle x^{2^{n-1}}}{\displaystyle 1-x^{2^n}} \)

5.\( \displaystyle \sum_{n=0}^{\infty}\dfrac{\displaystyle 2^n x^{x^{2^n}-1}}{\displaystyle 1+x^{2^n}} \)

 

[Markov2]

Write first series as $\displaystyle\sum\limits_{n=1}^{\infty }{\sum\limits_{j=1}^{n}{\frac{1}{{{2}^{n-1}}(j-1)!}}}=\sum\limits_{j=1}^{\infty }{\sum\limits_{n=j}^{\infty }{\frac{1}{{{2}^{n-1}}(j-1)!}}},$ we can reverse order of summation because of the positivity of the terms of the double series, so now things are straighforward.

You can do similar stuff on second series. Third series is easy by using the expansion of $e^x.$

 

[Sherlock1]

2. \( \displaystyle \sum_{n=1}^{\infty}\frac{1+2+2^2+\cdots+2^{n-1}}{n!}\)

$\displaystyle \sum_{n \ge 1}~\sum_{0 \le k \le n-1}\frac{2^k}{n!} = \sum_{n \ge 1}\frac{2^n-1}{n!} = \sum_{n \ge 0}\frac{2^n}{n!}-\sum_{n \ge 0}\frac{1}{n!} = e(e-1).$

 

[sbhatnagar]

1. \( \displaystyle \sum_{n=1}^{\infty}\frac{\left( 1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots +\dfrac{1}{(n-1)!} \right)}{2^{n-1}}\)

Solution to Problem 1:

Let \( \displaystyle S=\sum_{n=1}^{\infty}\frac{\left( 1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots +\dfrac{1}{(n-1)!} \right)}{2^{n-1}}\).

Let \( \displaystyle t_{n}=\frac{\left( 1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots +\dfrac{1}{(n-1)!} \right)}{2^{n-1}}\)

\( \displaystyle \begin{align*} t_{n+1} &= \frac{\left( 1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots +\dfrac{1}{(n-1)!}+\dfrac{1}{(n)!} \right)}{2^{n}} \\ t_{n+1} &= \frac{1}{2}\frac{\left( 1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots +\dfrac{1}{(n-1)!}+\dfrac{1}{(n)!} \right)}{2^{n-1}} \\ t_{n+1} &= \frac{1}{2}t_n +\frac{1}{2^n n!} \\ \sum_{n=0}^{\infty}t_{n+1} &=\frac{1}{2}\sum_{n=0}^{\infty}t_n + \sum_{n=0}^{\infty} \frac{1}{2^n n!} \\ S & =\frac{S}{2}+\sqrt{e} \\ S &= \boxed{2\sqrt{e}}\end{align*}\)

Try solving the last two problems. They are very tricky. :)

 

[Markov2]

My approach for first series is pretty short, the next step is $\displaystyle\sum\limits_{j = 1}^\infty {\frac{1}{{{2^{j - 1}}(j - 1)!}}\sum\limits_{n = 0}^\infty {\frac{1}{{{2^n}}}} } ,$ then combining the value of both series the result equals $2\sqrt e.$

 

[Sherlock1]

Try solving the last two problems. They are very tricky. :)

Don't post solutions yet. I'm working on them. (Nod)
Are you sure the last one is correct, though?


---------- Post added at 01:45 AM ---------- Previous post was at 12:00 AM ----------

$\displaystyle \begin{aligned} 3. ~f(x) & = \sum_{n \ge 1}\frac{(1+3^n)x^n}{n!} = \sum_{n \ge 1}\frac{x^n}{n!}+\sum_{n \ge 1}\frac{3^nx^n}{n!} \\& = e^{3x}+e^x-2 \Rightarrow f(\ln{3}) = 28.\end{aligned}$

$\displaystyle \begin{aligned} 5. ~ S & = \sum_{n \ge 0}\frac{2^nx^{2^n-1}}{1+x^{2^n}} = \sum_{n \ge 0}\frac{(1+x^{2^n})'}{1+x^{2^n}} = \bigg[\sum_{n \ge 0}\ln\left(1+x^{2^n}\right)\bigg]' \\& = \bigg[\ln\bigg(\prod_{n \ge 0} (1+x^{2^n})\bigg)\bigg]' = \bigg[\ln\bigg(\frac{1}{1-x}\bigg)\bigg]' =\boxed{\dfrac{1}{1-x}}. \end{aligned}$

 

[sbhatnagar]

Don't post solutions yet. I'm working on them. (Nod)
Are you sure the last one is correct, though?

Yes, you are correct. :)

 

[polygamma]

I wasn't aware that $\displaystyle \prod_{n=0}^{\infty} \left(1+x^{2^{n}} \right) = \sum_{n=0}^{\infty} x^{n} $. But from just writing out terms it appears to be true.

 

[Sherlock1]

I wasn't aware that $\displaystyle \prod_{n=0}^{\infty} \left(1+x^{2^{n}} \right) = \sum_{n=0}^{\infty} x^{n} $. But from just writing out terms it appears to be true.

We have $\displaystyle \prod_{0 \le k \le n}\left(1-x^{2^{k}}\right) = \frac{\left(1-x\right)}{\left(1-x^{2^{n+1}}\right)}\prod_{1 \le k \le n+1}\left(1-x^{2^{k}}\right) = \frac{\left(1-x\right) }{\left(1-x^{2^{n+1}}\right)}\prod_{0 \le k \le n}\left(1-x^{2^{k+1}}\right). $

But $\displaystyle \left(1-x^{2^{k+1}}\right) = \left(1-x^{2^{k}}\right)\left(1+x^{2^{k}}\right)$, thus $\displaystyle \prod_{0 \le k \le n}\left(1-x^{2^{k+1}}\right) =\prod_{0 \le k \le n}\left(1-x^{2^{k}}\right)\prod_{0 \le k \le n}\left(1+x^{2^{k}}\right). $

Thus $\displaystyle \frac{\left(1-x\right)}{\left(1-x^{2^{n}}\right)}\prod_{0 \le k \le n}\left(1+x^{2^{k}}\right) = \frac{\prod_{0 \le k \le n}\left(1-x^{2^{k}}\right)}{\prod_{0 \le k \le n}\left(1-x^{2^{k}}\right)} = 1 \implies \prod_{0 \le k \le n}\left(1+x^{2^{k}}\right) = \frac{1-x^{2^{n+1}}}{1-x}. $

The difference of two squares can do wonders! We get our case when $n \to \infty$ of course.

 

[sbhatnagar]

4. \( \displaystyle \sum_{n=1}^{\infty}\dfrac{\displaystyle x^{2^{n-1}}}{\displaystyle 1-x^{2^n}} \)

Let \( \displaystyle S=\sum_{n=1}^{n}\dfrac{\displaystyle x^{2^{n-1}}}{\displaystyle 1-x^{2^n}} \)

\( \displaystyle S=\sum_{n=1}^{\infty}\dfrac{\displaystyle x^{2^{n-1}}}{\displaystyle 1-x^{2^n}} =\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}+1-1}{(1-x^{2^{n-1}})(1+x^{2^{n-1}})} = \sum_{n=1}^{\infty} \frac{1}{(1-x^{2^{n-1}})}-\frac{1}{(1-x^{2^{n-1}})(1+x^{2^{n-1}})} = \sum_{n=1}^{\infty}\frac{1}{(1-x^{2^{n-1}})}-\frac{1}{(1-x^{2^{n}})}\)

\( \displaystyle S=\frac{1}{1-x}-\frac{1}{1-x^2}+\frac{1}{1-x^2}-\frac{1}{1-x^4}+\cdots= \boxed{\dfrac{1}{1-x}}\)

(Music)

 

[melese]

We have $\displaystyle \prod_{0 \le k \le n}\left(1-x^{2^{k}}\right) = \frac{\left(1-x\right)}{\left(1-x^{2^{n+1}}\right)}\prod_{1 \le k \le n+1}\left(1-x^{2^{k}}\right) = \frac{\left(1-x\right) }{\left(1-x^{2^{n+1}}\right)}\prod_{0 \le k \le n}\left(1-x^{2^{k+1}}\right). $

But $\displaystyle \left(1-x^{2^{k+1}}\right) = \left(1-x^{2^{k}}\right)\left(1+x^{2^{k}}\right)$, thus $\displaystyle \prod_{0 \le k \le n}\left(1-x^{2^{k+1}}\right) =\prod_{0 \le k \le n}\left(1-x^{2^{k}}\right)\prod_{0 \le k \le n}\left(1+x^{2^{k}}\right). $

Thus $\displaystyle \frac{\left(1-x\right)}{\left(1-x^{2^{n}}\right)}\prod_{0 \le k \le n}\left(1+x^{2^{k}}\right) = \frac{\prod_{0 \le k \le n}\left(1-x^{2^{k}}\right)}{\prod_{0 \le k \le n}\left(1-x^{2^{k}}\right)} = 1 \implies \prod_{0 \le k \le n}\left(1+x^{2^{k}}\right) = \frac{1-x^{2^{n+1}}}{1-x}. $

The difference of two squares can do wonders! We get our case when $n \to \infty$ of course.

I think there's a simpler reason. When multiplying out, we find that $\displaystyle\prod_{0\leq k\leq n}(1+x^{2^k})$ is the sum of terms of the form $x^m$ (except for $1$), where $m$ is a sum of powers of two. Then we may notice that any $1\leq m\leq2^{n+1}-1$ is obtained exactly once because any number can be written uniquely in base-2 representation.

For example, take $n=3$: $(1+x^{2^0})(1+x^{2^1})(1+x^{2^2})(1+x^{2^3})=1+x^{2^0}+x^{2^1}+x^{2^0+2^1}+x^{2^2}+x^{2^0+2^2}+x^{2^1+2^2}+x^{2^0+2^2+2^2}=1+x+x^2+x^3+x^4+x^5+x^6+x^7$

Therefore, $\displaystyle\prod_{0\leq k\leq n}(1+x^{2^k})=1+\sum_{1\leq m\leq 2^{n+1}-1}x^m$

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