# Finding the sum of a series

I have the following series that I came up with in doing a problem: ##\displaystyle \sum_{n=0}^{\infty} \frac{1}{2^{n+1}(n+1)}##. I looked at WolframAlpha and it says that this series converges to ##\log (2)##. Is it possible to figure this out analytically?

fresh_42
Mentor
2021 Award
I have the following series that I came up with in doing a problem: ##\displaystyle \sum_{n=0}^{\infty} \frac{1}{2^{n+1}(n+1)}##. I looked at WolframAlpha and it says that this series converges to ##\log (2)##. Is it possible to figure this out analytically?
As the result involves the logarithm, we first have to find a way to express the logarithm by a series. Which one do you chose? If you define the logarithm otherwise, there will probably a bit more work to do.

As the result involves the logarithm, we first have to find a way to express the logarithm by a series. Which one do you chose? If you define the logarithm otherwise, there will probably a bit more work to do.
Well, suppose that I had no idea that the answer turned out to be ##\log 2##. Would there be any way for me to proceed?

StoneTemplePython
Gold Member
Pattern recognition and a slightly more general problem:

use ## p \in (0,1)## instead of ##\frac{1}{2}##

what happens if you differentiate the series?

fresh_42
Mentor
2021 Award
Well, suppose that I had no idea that the answer turned out to be ##\log 2##. Would there be any way for me to proceed?
The easiest way is to consider ##\log(1+x)=-\sum_{k=0}^{\infty}\dfrac{(-x)^{k+1}}{k+1}## which is the power series expansion at ##x=0## for ##-1< x \le 1\,.## Then simply take ##x=-\frac{1}{2}\,.## But I think it's fun (and a good exercise) to consider the other proposed ways above, esp. what @StoneTemplePython suggested: differentiate ##f(p)=\sum_{k=0}^{\infty}\dfrac{p^{k+1}}{k+1}##.

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• mfb
Pattern recognition and a slightly more general problem:

use ## p \in (0,1)## instead of ##\frac{1}{2}##

what happens if you differentiate the series?
That's neat... So if I let ##S## be the desired sum, I found that ##S(p) = \log p + C##. How would I find the constant of integration? I don't know of any initial conditions where ##0 < p < 1##.

StoneTemplePython
Gold Member
That's neat... So if I let ##S## be the desired sum, I found that ##S(p) = \log p + C##. How would I find the constant of integration? I don't know of any initial conditions where ##0 < p < 1##.

actually you do know the initial conditions are

##
\displaystyle \sum_{n=0}^{\infty} \frac{p^{n+1}}{(n+1)}
##

so when you have

##\frac{1}{1-p} = 1 + p + p^2 + p^3 + ... ##

you then integrate both sides... ignoring the constant of integration for a moment, does the term by term integration of the Right hand side match with your original series? If so, the Left hand side does too -- this allows you to "zero in" on the constant of integration.

- - - -
edit: another simpler thing to notice is that every term in your original series has a ##p## there are no constants...

another edit: changed the original series to make sure the ##p## is in the numerator!

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fresh_42
Mentor
2021 Award
That's neat... So if I let ##S## be the desired sum, I found that ##S(p) = \log p + C##. How would I find the constant of integration? I don't know of any initial conditions where ##0 < p < 1##.
What is ##\sum q^k## for ##0<q<1\,##?

actually you do know the initial conditions are

##
\displaystyle \sum_{n=0}^{\infty} \frac{1}{p^{n+1}(n+1)}
##

so when you have

##\frac{1}{1-p} = 1 + p + p^2 + p^3 + ... ##

you then integrate both sides... ignoring the constant of integration for a moment, does the term by term integration of the Right hand side match with your original series? If so, the Left hand side does too -- this allows you to "zero in" on the constant of integration.

- - - -
edit: another simpler thing to notice is that every term in your original series has a ##p## there are no constants...
So are you saying that the constant of integration is 0?

StoneTemplePython
Gold Member
So are you saying that the constant of integration is 0?

You tell me. When you applied the derivative operator to your original series, did any term get mapped to zero? That is the information loss associated with basic differentiation. So what kind of information was lost here?

You tell me. When you applied the derivative operator to your original series, did any term get mapped to zero? That is the information loss associated with basic differentiation. So what kind of information was lost here?
I would suppose none... But if ##S(p) = \log p## then ##S(1/2) = -\log 2## which has a sign error

StoneTemplePython
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