# Infinite Series: $\sum_{n=1}^\infty \frac{1}{2^nn}$

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!

• Hint: What function has a Taylor series resembling $\sum x^n/n$? – Moya Apr 8 '16 at 4:07
• e^x is close to that with $\sum x^n/n!$. – Grant Stenger Apr 8 '16 at 4:10
• Hint: what if you differentiated $\sum x^n/n$? What series do you get? – User8128 Apr 8 '16 at 4:11
• sum x^(n-1) right? – Grant Stenger Apr 8 '16 at 4:12
• 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! – Grant Stenger 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 $a \in [0,1)$

• 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. – Grant Stenger Apr 8 '16 at 4:20
• In this case the series is $e^{1/2}$. You have to know the Taylor series for some basic functions. – kmitov Apr 8 '16 at 4:26
• Also, substituting x=1/2 means the sum = -ln(1/2), which according to WolframAlpha should be ln(2). Thoughts? – Grant Stenger Apr 8 '16 at 4:26
• I wrote the same. – kmitov Apr 8 '16 at 4:27
• How did you get ln(2)? – Grant Stenger Apr 8 '16 at 4:30