Sum to Infinity: Proving Divergence

[yecko]

Homework Statement

evaluate $\sum _{n=1}^{\infty }4^{\frac{1}{n}}-4^{\frac{1}{n+2}}$ .

https://holland.pk/uptow/i4/fc981b864d95a636c4f08b9deb209cd6.png

Homework Equations

telescoping series: sum = infinite lim (a1-a(n+1))
S=a/(1-r)

The Attempt at a Solution

as the latter function is of "n-2" instead of "n-1", it isn't telescoping series, right?
however, I don't know how to prove it is divergent instead of convergent in this question.
thank you very much!

 

[Ray Vickson]

Homework Statement

evaluate $\sum _{n=1}^{\infty }4^{\frac{1}{n}}-4^{\frac{1}{n+2}}$ .
View attachment 203669
https://holland.pk/uptow/i4/fc981b864d95a636c4f08b9deb209cd6.png

Homework Equations

telescoping series: sum = infinite lim (a1-a(n+1))
S=a/(1-r)

The Attempt at a Solution

as the latter function is of "n-2" instead of "n-1", it isn't telescoping series, right?
however, I don't know how to prove it is divergent instead of convergent in this question.
thank you very much!

Why do you think it is divergent? Are you sure that Wolfram Alpha has not gotten it wrong?

Try a few finite sums first, such as $\sum_{n=1}^4 \cdots$ or $\sum_{n=1}^8 \cdots$. This will give you some valuable insights.

 

[yecko]

Indeed, it is one of the past exam paper question which we are not allowed to use calculator.
To calculate by hand would be far too time consuming in exam.

∑4n=1⋯∑n=14⋯\sum_{n=1}^4 \cdots or ∑8n=1⋯∑n=18⋯\sum_{n=1}^8 \cdots.

Other than try out the values, is there any other ways to calculate it?
Thanks.

 

[DrClaude]

Strangely, I don't get the same answer as you do with exactly the same input on WolframAlpha.

In any case, you should use parentheses, as in your case WolframAlpha considered only the first term as part of the sum.

 

[yecko]

Strangely, I don't get the same answer as you do with exactly the same input on WolframAlpha.

In any case, you should use parentheses, as in your case WolframAlpha considered only the first term as part of the sum.

Thanks for telling me that. The new answer I got from WolframAlpha is 3.98619, about 4.
How can I get it by hand?
Thanks

 

[Ray Vickson]

Thanks for telling me that. The new answer I got is 3.98619, about 4.
How can I get it by hand?
Thanks

You can do what I suggested in post #2. I never suggested doing calculations; I suggested writing down the sums, and all that is perfectly feasible in an exam without use of a calculator.

Anyway, the Wolfram Alpha result is close, but not exact.

 

[yecko]

But I need to have steps in long question... What I mean is an accurate answer in each steps...
Calculating by trials isn't kind of valid steps, isn't it?

 

[Ray Vickson]

But I need to have steps in long question... What I mean is an accurate answer in each steps...
Calculating by trials isn't kind of valid steps, isn't it?

Instead of arguing about it, why don't you just try what I suggested? I did not make the suggestion lightly, just to waste your time; I had a very good reason.

 

[Buffu]

@Ray Vickson
Is $3.9$ or $6$ the answer ?

 

[WWGD]

Like Ray Vickson suggested, rewrite the sum as a "sum of sums" and change the indices so that there is some cancellation.

 

[Ray Vickson]

@Ray Vickson
Is $3.9$ or $6$ the answer ?

No, neither is correct.

Have you done what I suggested in post #2?

 

[Buffu]

No, neither is correct.

Have you done what I suggested in post #2?

Yes, I did that.

I found, $\displaystyle \sum^\infty_{n = 1} 4^{1/n} = 4 + 4^{1/2} + 4^{1/3} + ...$ And $\displaystyle \sum^\infty_{n = 1} 4^{1/(n+2)} = 4^{1/3} + 4^{1/4} + ...$

So the red part in $\displaystyle \sum^\infty_{n = 1} 4^{1/n} = 4 + 4^{1/2} + \color{red}{4^{1/3} + 4^{1/4} + ... }$ is $\displaystyle \sum^\infty_{n = 1} 4^{1/(n+2)}$.

$\displaystyle \sum^\infty_{n = 1} 4^{1/n} = 4 + 4^{1/2} + \color{red}{4^{1/3} + 4^{1/4} + ... } = \sum^\infty_{n = 1} 4^{1/n} = 4 + 4^{1/2} + \sum^\infty_{n = 1} 4^{1/(n+2)} \iff \\ \displaystyle \sum^\infty_{n = 1} 4^{1/n} - \sum^\infty_{n = 1} 4^{1/(n+2)} = 4 + 2 = 6$

Why this wrong ?

 

[Mark44]

Yes, I did that.

I found, $\displaystyle \sum^\infty_{n = 1} 4^{1/n} = 4 + 4^{1/2} + 4^{1/3} + ...$ And $\displaystyle \sum^\infty_{n = 1} 4^{1/(n+2)} = 4^{1/3} + 4^{1/4} + ...$

So the red part in $\displaystyle \sum^\infty_{n = 1} 4^{1/n} = 4 + 4^{1/2} + \color{red}{4^{1/3} + 4^{1/4} + ... }$ is $\displaystyle \sum^\infty_{n = 1} 4^{1/(n+2)}$.

$\displaystyle \sum^\infty_{n = 1} 4^{1/n} = 4 + 4^{1/2} + \color{red}{4^{1/3} + 4^{1/4} + ... } = \sum^\infty_{n = 1} 4^{1/n} = 4 + 4^{1/2} + \sum^\infty_{n = 1} 4^{1/(n+2)} \iff \\ \displaystyle \sum^\infty_{n = 1} 4^{1/n} - \sum^\infty_{n = 1} 4^{1/(n+2)} = 4 + 2 = 6$

Why this wrong ?

Look at the partial sum, $\sum_{n=1}^k \left( 4^{1/k} - 4^{\frac 1 {k+2}}\right)$. I find it helpful to write the positive terms in a column on the left and the terms being subtracted in a column on the right. There is a lot of telescoping going on, but possibly not as much as you think.

Be sure to write the terms for k=1 through 6 or so, as well as the last 4 or 5 terms. Then take the limit as $k \to \infty$. Remember that the sum of an infinite series is the limit of the sequence of partial sums as more and more terms are taken.

 

[Mark44]

This is what I'm describing in the previous post.
$4^1~~~ -4^{1/3}$
$4^{1/2}~~~ -4^{1/4}$
$4^{1/3}~~~ -4^{1/5}$
$4^{1/4}~~~ -4^{1/6}$
$4^{1/5}~~~ -4^{1/7}$
$\vdots$
Be sure to continue the pattern for the last three or four pairs of terms.

 

[Buffu]

This is what I'm describing in the previous post.
$4^1~~~ -4^{1/3}$
$4^{1/2}~~~ -4^{1/4}$
$4^{1/3}~~~ -4^{1/5}$
$4^{1/4}~~~ -4^{1/6}$
$4^{1/5}~~~ -4^{1/7}$
$\vdots$
Be sure to continue the pattern for the last three or four pairs of terms.

$\\ 4^{1/1} - 4^{1/3}\\ 4^{1/2} - 4^{1/4}\\ 4^{1/3} - 4^{1/5}\\~~~ \vdots \\ 4^{1/(n-2)} - 4^{1/(n)}\\ 4^{1/(n-1)} - 4^{1/(n+1)}\\ 4^{1/(n)} - 4^{1/(n+2)}\\$

So sum to $n$ is $4 + 4^{1/2} - 4^{1/(n+1)} - 4^{1/(n+2)}$ Now I take limit n to infinity ??

 

[Mark44]

$\\ 4^{1/1} - 4^{1/3}\\ 4^{1/2} - 4^{1/4}\\ 4^{1/3} - 4^{1/5}\\~~~ \vdots \\ 4^{1/(n-2)} - 4^{1/(n)}\\ 4^{1/(n-1)} - 4^{1/(n+1)}\\ 4^{1/(n)} - 4^{1/(n+2)}\\$

So sum to $n$ is $4 + 4^{1/2} - 4^{1/(n+1)} - 4^{1/(n+2)}$ Now I take limit n to infinity.

Yes, and the limit as $n \to \infty$ is what?

 

[Buffu]

Yes, and the limit as $n \to \infty$ is what?

0 ?

 

[Mark44]

0 ?

No, but this is Yecko's problem. Don't do his work for him.

 

[Buffu]

No, but this is Yecko's problem. Don't do his work for him.

Yes got it !

 

[Ray Vickson]

This is what I'm describing in the previous post.
$4^1~~~ -4^{1/3}$
$4^{1/2}~~~ -4^{1/4}$
$4^{1/3}~~~ -4^{1/5}$
$4^{1/4}~~~ -4^{1/6}$
$4^{1/5}~~~ -4^{1/7}$
$\vdots$
Be sure to continue the pattern for the last three or four pairs of terms.

That is exactly what I suggested the OP do, but he refused to cooperate, so I gave up.

 

[Mark44]

That is exactly what I suggested the OP do, but he refused to cooperate, so I gave up.

Well, you can lead a horse to water...

 

[yecko]

Sorry guys~
I was not reading this forum yesterday night because I have another exam...

Yes, I did that.

I found, ∞∑n=141/n=4+41/2+41/3+...∑n=1∞41/n=4+41/2+41/3+...\displaystyle \sum^\infty_{n = 1} 4^{1/n} = 4 + 4^{1/2} + 4^{1/3} + ... And ∞∑n=141/(n+2)=41/3+41/4+...∑n=1∞41/(n+2)=41/3+41/4+...\displaystyle \sum^\infty_{n = 1} 4^{1/(n+2)} = 4^{1/3} + 4^{1/4} + ...

So the red part in ∞∑n=141/n=4+41/2+41/3+41/4+...∑n=1∞41/n=4+41/2+41/3+41/4+...\displaystyle \sum^\infty_{n = 1} 4^{1/n} = 4 + 4^{1/2} + \color{red}{4^{1/3} + 4^{1/4} + ... } is ∞∑n=141/(n+2)∑n=1∞41/(n+2)\displaystyle \sum^\infty_{n = 1} 4^{1/(n+2)} .

∞∑n=141/n=4+41/2+41/3+41/4+...=∞∑n=141/n=4+41/2+∞∑n=141/(n+2)⟺∞∑n=141/n−∞∑n=141/(n+2)=4+2=6

6 should be the correct answer, as I worked out just now, where I agree with Buffu...
Also, there is no "0" as the option from any version of the answer... (though there are numerous typo and problems on them...)

I worked this way: by telescoping series,
$\sum _{n=1}^{\infty }\left(4^{\frac{1}{n}}-4^{\frac{1}{n+1}}+4^{\frac{1}{n+1}}-4^{\frac{1}{n+2}}\right)=\lim _{x\to \infty }\left(4-4^{\frac{1}{n+1}}+4^{\frac{1}{2}}-4^{\frac{1}{n+2}}\right)=4-0+2-0=6$

Do you guys agree with this approach? seems the same steps as you guys??
(still not the same answer as wolfram alpha suggested:4!)

moreover, I still don't see the reason to do this... I just don't get it...

41/1−41/341/2−41/441/3−41/5 ⋮41/(n−2)−41/(n)41/(n−1)−41/(n+1)41/(n)−41/(n+2)41/1−41/341/2−41/441/3−41/5 ⋮41/(n−2)−41/(n)41/(n−1)−41/(n+1)41/(n)−41/(n+2)\\ 4^{1/1} - 4^{1/3}\\ 4^{1/2} - 4^{1/4}\\ 4^{1/3} - 4^{1/5}\\~~~ \vdots \\ 4^{1/(n-2)} - 4^{1/(n)}\\ 4^{1/(n-1)} - 4^{1/(n+1)}\\ 4^{1/(n)} - 4^{1/(n+2)}\\

 

[Buffu]

Sorry guys~
I was not reading this forum yesterday night because I have another exam...

6 should be the correct answer, as I worked out just now, where I agree with Buffu...
Also, there is no "0" as the option from any version of the answer... (though there are numerous typo and problems on them...)

I worked this way: by telescoping series,
$\sum _{n=1}^{\infty }\left(4^{\frac{1}{n}}-4^{\frac{1}{n+1}}+4^{\frac{1}{n+1}}-4^{\frac{1}{n+2}}\right)=\lim _{x\to \infty }\left(4-4^{\frac{1}{n+1}}+4^{\frac{1}{2}}-4^{\frac{1}{n+2}}\right)=4-0+2-0=6$

Do you guys agree with this approach? seems the same steps as you guys??
(still not the same answer as wolfram alpha suggested:4!)

moreover, I still don't see the reason to do this... I just don't get it...

6 is wrong since

$x^0 = 1$ not 0.

I did the same mistake twice lol.

 

[Mark44]

6 should be the correct answer, as I worked out just now, where I agree with Buffu...
Also, there is no "0" as the option from any version of the answer... (though there are numerous typo and problems on them...)

Look at posts #14 and #15.