# Divergence of a modified alternating series

I'm currently trying to prove that $\sum_{k=1}^\infty t_k$ diverges, where $$t_k = \frac{(-1)^k+1}{2k}+ \frac{(-1)^k}{\sqrt{k}}$$

I know that $\sum_{k=1}^\infty \frac{(-1)^k}{\sqrt{k}}$ converges by the alternating series test, and that the series $\sum_{k=1}^\infty \frac{(-1)^k+1}{2k}$ diverges, as it's just a scaled harmonic series interspersed with zeros.

My question is: is it correct to say that the series in question diverges because it is the sum of a convergent series and a divergent series? It seems pretty straightforward if this is allowed, but I'm worried that I'm missing some subtlety about saying $\sum t_k = \sum_{k=1}^\infty \frac{(-1)^k}{\sqrt{k}} +\sum_{k=1}^\infty \frac{(-1)^k+1}{2k}$, (i.e., rearranging the terms in the sum), since the convergent part is only conditionally so.

Would someone please be able to clarify if such a subtlety exists, or if what I've done is alright? Thanks!

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You are right to ne cautious. But look at the sum $S_n$ of the first $n$ terms, $n$ large. The part due to $1/2+1/4+\cdots$ is (kind of) big, while the part due to the $\sum_1^n (-1)^n$ doesn't get very negative, indeed stays abovve $-1$. So $S_n$ is kind of big. –  André Nicolas Feb 7 '13 at 22:51

If $\displaystyle{\sum_{k=1}^\infty a_k=a, \ \sum_{k=1}^\infty b_k=b}$ then $\displaystyle{\sum_{k=1}^\infty (a_k+b_k)=a+b}$.

Now if your series, $\displaystyle{\sum_{k=1}^\infty t_k}$, converges then the series $\displaystyle{\sum_{k=1}^\infty t_k+s_k}$, where $s_k=\dfrac{(-1)^{k+1}}{\sqrt{k}}$, converges ↯.

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That helped, thanks! –  sourisse Feb 7 '13 at 21:54

Hint:

Write your general term as $$t_k=a_k+b_k+c_k$$ in such a way that $\sum a_k$ and $\sum b_k$ converge by Leibniz criterion for alternating series, and $\sum c_k$ diverges because it is a constant times the harmonic series.

If $\sum t_k$ converged, then so would $\sum t_k-a_k-b_k=\sum c_k$. A contradiction.