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?


  • $\begingroup$ 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 $\sum x^n/n!$. $\endgroup$ Apr 8 '16 at 4:10
  • 1
    $\begingroup$ 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? $\endgroup$ Apr 8 '16 at 4:12
  • 1
    $\begingroup$ 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! $\endgroup$ Apr 8 '16 at 4:18

$\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 $a \in [0,1)$

  • $\begingroup$ 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. $\endgroup$ 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? $\endgroup$ 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)? $\endgroup$ Apr 8 '16 at 4:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.