# Convergence of $\sum_{n=2}^\infty{(-1)^n\over{\ln{n}}}{\left({1\over2}+{2\over2^2}+\cdots+{n\over2^n}\right)}$

$$\sum_{n=2}^\infty{(-1)^n\over{\ln{n}}}{\left({1\over2}+{2\over2^2}+\cdots+{n\over2^n}\right)}$$

I've used the comparison test with $n^2\over{\ln{n}2^n}$ which converges with the root test to prove absolute convergence of the series. Do I have a mistake?

• did you mean to write 1/2 + 2/2^2 + ... + n/2^n ? – Adam Rubinson Jan 30 '13 at 8:46
• There was a typo, thanks – Gyt Jan 30 '13 at 8:50
• This series is not absolutely convergent. – Olivier Bégassat Jan 30 '13 at 8:51
• Babak - why do you say that? That's not obvious to me. Another thing to notice is that if the (-1) were changed to 1 then clearly the sum would diverge to +infinity: just think about very large n. – Adam Rubinson Jan 30 '13 at 8:55
• It does not converge absolutely. The terms have absolute value bigger than $\frac{1}{2\log n}$, which goes to $0$ much too slowly, – André Nicolas Jan 30 '13 at 8:56

$$\sum_{k=1}^n \dfrac{k}{2^k} = 2 - \dfrac{n+2}{2^n}$$ This can be derived as follows. We have that $$\sum_{k=1}^n x^k = \dfrac{x-x^{n+1}}{1-x}$$ Differentiate to get that $$\sum_{k=1}^n k x^{k-1} = \dfrac{nx^{n+1} - (n+1)x^n+1}{(1-x)^2} \implies \sum_{k=1}^n k x^{k} = \dfrac{nx^{n+2} - (n+1)x^{n+1}+x}{(1-x)^2}$$ Take $x=1/2$ to get what we want. Hence, your sum is $$\sum_{n=1}^{\infty}(-1)^n \dfrac1{\ln(n)} \left(2 - \dfrac{n+2}{2^n}\right)$$ Now use alternate series test to conclude the series converges.
The series doesn't converge absolutely. To see this, note that $$\left\vert (-1)^n \dfrac1{\ln(n)} \left(2 - \dfrac{n+2}{2^n}\right)\right\vert = \dfrac1{\ln(n)}\left(2 - \dfrac{n+2}{2^n}\right) \geq \dfrac1{2 \ln(n)}$$ and we know that $$\sum_{n=1}^{\infty}\dfrac1{\ln(n)}$$ diverges.