Prove that $\sum_{n=1}^{\infty}(-1)^n\frac{1}{4^n}\frac{[3+(-1)^n]^n}{n}$ diverges.

Question

I've tried writing it as a sum of two series, such as $\sum_{n=1}^{\infty}(-1)^n\frac{1}{4^n}\frac{[3+(-1)^n]^n}{n}=\sum_{n=0}^{\infty}(-1)^{2n+1}\frac{1}{2^{2n+1}(2n+1)}+\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n}$ and I thought that the divergence of the second sum implies divergence of the original one. But I'm not sure under what circumstances this method is allowed, because I've changed the order of original terms to get those two sums. I'll be thankful for any idea how to prove the original task.

Answer 1

Your method is fine indeed

• for $n=2k \implies (-1)^n\frac{1}{4^n}\frac{[3+(-1)^n]^n}{n}=\frac{1}{2k}$
• for $n=2k-1 \implies (-1)^n\frac{1}{4^n}\frac{[3+(-1)^n]^n}{n}=-\frac{1}{2^{2k-1}(2k-1)}$

and therefore

$$\sum_{n=1}^{2N}(-1)^n\frac{1}{4^n}\frac{[3+(-1)^n]^n}{n}=\frac12\sum_{k=1}^{N}\frac1k-\sum_{k=1}^{N}\frac{1}{2^{2k-1}(2k-1)} \to \infty$$

Refer also to the related

• If an infinite series is divergent, after rearrangement can it be convergent?
• Can a divergent alternating series by rearrangement of terms be made to converge to a value?
• Sum of Divergent and Convergent Series

Answer 2

The re arrangement argument is not valid in general, refer Riemann re arrangement theorem. In this case however, we can work around this. This is because the first series is convergent.

Therefore we can get $$S_{2k}\geq \frac{1}{2}\sum_{i=1}^k \frac{1}{i} - c$$ for some fixed constant $c$ (where $S_{2k}$ denotes the partial sum upto $2k$ indices). Therefore we can see that $S_{2k}\to\infty$.