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I am having troubles to explain if the following equality holds or not $$\lim_{k\to\infty}\sum_{n=1}^{\infty}(-1)^{n}(n+k)^{-1}=\sum_{n=1}^{\infty}(-1)^{n}\lim_{k\to\infty}(n+k)^{-1}=0.$$ As far as I see, I can't apply the dominated convergence theorem since $|f_{k}(n)|=|(-1)^{n}(n+k)^{-1}|=(n+k)^{-1}$ can't be dominated by summable sequence over $n$. How could I proceed?

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  • $\begingroup$ The first of your equalities comes unjustified. But with a conditional convergence, you can easily apply whatever-you-want to, say, pairwise summation. $\endgroup$ – metamorphy Sep 1 '18 at 17:04
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HINT:

Note that we can write

$$\sum_{n=1}^\infty \frac{(-1)^n}{n+k}=\sum_{n=1}^\infty\left(\frac{1}{2n+k}-\frac{1}{2n-1+k}\right)$$

Can you proceed now?

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Or: $$ \sum_{n\geq 1}\frac{(-1)^n}{n+k} = \sum_{n\geq 1}\int_{0}^{+\infty}(-1)^n e^{-nx} e^{-kx}\,dx =\int_{0}^{+\infty}\frac{dx}{(1+e^x)e^{kx}}\leq \int_{0}^{+\infty}\frac{dx}{2e^{kx}}=\frac{1}{2k}.$$

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