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Just encountered a problem in my BC Calculus sequences and series unit that I just can't figure out.

I don't know the latex, but the problem was to find compute $$ \sum_{n=1}^\infty \frac{1}{2^nn} $$ The handout was just on convergence, so it was easy to see that by ratio test, it does converge, but I think I'm missing some ideas to be able to calculate the sum.

We have done Taylor series, I just don't quite know how to go backwards, although my guess is that you find a way to write the partial sums and find it's value as the limit approaches infinity.

How would you go about solving this?

Thanks!

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  • $\begingroup$ Hint: What function has a Taylor series resembling $\sum x^n/n$? $\endgroup$ – Moya Apr 8 '16 at 4:07
  • $\begingroup$ e^x is close to that with $\sum x^n/n!$. $\endgroup$ – Grant Stenger Apr 8 '16 at 4:10
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    $\begingroup$ Hint: what if you differentiated $\sum x^n/n$? What series do you get? $\endgroup$ – User8128 Apr 8 '16 at 4:11
  • $\begingroup$ sum x^(n-1) right? $\endgroup$ – Grant Stenger Apr 8 '16 at 4:12
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    $\begingroup$ Ahh, got it. It's the sum of x^(n-1), which can easily be computed, so just integrate 1/(1-x) and make x=1/2. That makes a lot of sense, thank you so much! $\endgroup$ – Grant Stenger Apr 8 '16 at 4:18
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$\sum_{n=0}^\infty x^n = \frac{1}{1-x}, $ for $|x|<1$. Geometric series.

Integrate both sides to get

$\sum_{n=1}^\infty \frac{x^n}{n} = C-\ln (1-x), $ for $|x|<1$.

For $x=0$ you see that $C=0$. Then substitute $x=1/2$ to get $\ln 2$

The integration is valid for any closed interval $[-a,a]$ for $a \in [0,1)$

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  • $\begingroup$ That makes a lot of sense thank you so much. As a follow up, the sum of 1/(2^n*n!) was also asked on this handout. What would be your approach? Integrals seem pretty tricky with factorials. $\endgroup$ – Grant Stenger Apr 8 '16 at 4:20
  • $\begingroup$ In this case the series is $e^{1/2}$. You have to know the Taylor series for some basic functions. $\endgroup$ – kmitov Apr 8 '16 at 4:26
  • $\begingroup$ Also, substituting x=1/2 means the sum = -ln(1/2), which according to WolframAlpha should be ln(2). Thoughts? $\endgroup$ – Grant Stenger Apr 8 '16 at 4:26
  • $\begingroup$ I wrote the same. $\endgroup$ – kmitov Apr 8 '16 at 4:27
  • $\begingroup$ How did you get ln(2)? $\endgroup$ – Grant Stenger Apr 8 '16 at 4:30

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