I am given the following sum:
$$\sum_{n=0}^\infty{\frac{1}{2^{n+2}(n+2)}}$$
How do I find what is it equal to? It sort of looks like a logarithm, but not exactly, so how do I proceed to solve this problem?
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Sartr
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3Does this look better to you? $$\sum_{n=0}^{\infty} \frac{x^{n+2}}{n+2}$$ – Jean-Claude Arbaut Apr 21 '17 at 18:22
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2Or is $$-x+\sum_{n=1}^\infty\frac{x^n}n$$ even better? – Angina Seng Apr 21 '17 at 18:24
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How did you reach $$-x+\sum_{n=1}^\infty\frac{x^n}n$$? Can you post an explanation on how you reached it? – Sartr Apr 21 '17 at 18:31
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1I changed the variable in Jean-Claude's formula :-) – Angina Seng Apr 21 '17 at 18:32
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How so? I'm a bit slow at understanding... – Sartr Apr 21 '17 at 18:43
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For example if I had $$\sum_{n=0}^\infty{\frac1{2^{n+4}(n+4)}}$$ how would I proceed – Sartr Apr 21 '17 at 18:51
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@Sartr See my HINT posted herein. – Mark Viola Apr 21 '17 at 18:51
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Yeah, I saw it, but still don't get it :) – Sartr Apr 21 '17 at 18:55
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@Sartr I've edited to add a "Spoiler Alert." – Mark Viola Apr 21 '17 at 19:38
2 Answers
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HINT:
Note that
$$\int_0^x t^{n+1}\,dt=\frac{x^{n+2}}{n+2} \tag 1$$
Now, sum both sides over $n$ and let $x=1/2$.
SPOILER ALERT: Scroll over the highlighted area to reveal the solution
For $|x|<1$, the series $\sum_{n=0}^\infty \frac{x^{n+2}}{n+2}$ converges. So, for $|x|<1$, we sum both sides of $(1)$ to obtain $$\sum_{n=0}^\infty \frac{x^{n+2}}{n+2}=\sum_{n=0}^\infty \int_0^x t^{n+1}\,dt \tag 2$$For all $x_0<1$ and $|x|\le x_0$, the uniform convergence of the series $\sum_{n=0}^\infty t^n$ allows us to interchange the summation and the integration in $(2)$. Proceeding, we find that $$\begin{align}\\\sum_{n=0}^\infty \frac{x^{n+2}}{n+2}&=\int_0^x \sum_{n=0}^\infty t^{n+1}\,dt\\\\&=\int_0^x \frac{t}{1-t}\,dt\\\\&=-x-\log(1-x)\tag 3\\\end{align}$$ Letting $x=1/2$ in $(3)$ yields the coveted result $$\sum_{n=0}^\infty \frac{1}{(n+2)2^{n+2}}=\log(2)-\frac12$$And we are done!
Mark Viola
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You can interchange the order of integration and the series. The series is a geometric one $\sum_{n=0}^\infty t^{n+1}$. Then, integrate the result. – Mark Viola Apr 21 '17 at 18:47
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Woah, that makes sense, it's a shame they missed to teach us that. Thank you a lot! – Sartr Apr 21 '17 at 21:59
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1You're quite welcome. My pleasure and pleased this was helpful. -Mark – Mark Viola Apr 21 '17 at 22:07
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If you know that
$$-\log(1-x)=\sum_{k=1}^\infty\frac{x^k}k$$
you can set $x=\frac12$ and deduce the first term. Then
$$\frac12+\sum_{k=2}^\infty\frac1{2^kk}=\log2.$$