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Exercise: Suppose that $a_k \geq 0$ for $k$ large and that $\sum_{k = 1}^{\infty} \frac{a_k}{k}$ converges.

Prove that $$\lim_{j \to \infty}\sum_{k = 1}^{\infty} \frac{a_k}{j+k} = 0$$

Attempt in proof:

Suppose $a_k \geq 0$ for any large $k$ and that $\sum_{k = 1}^{\infty} \frac{a_k}{k}$ converges. . Then give $\epsilon >0$ there is $N \in N$ such that $\left|\sum_{k = n}^{\infty} \frac{a_k}{k}\right| < \epsilon$.

Let $s_n = \frac{a_n}{j+k}$ denote the partial sums.

Then $\sum_{k = 1}^{\infty} \frac{a_k}{j+k} $ will converge to zero if and only if its partial sum converges to zero as n approaches infinity.

Taking the limit of $$\lim_{j \to \infty}\sum_{k = 1}^{\infty} \frac{a_k}{j+k} = \lim_{j \to \infty}\frac{a_1}{j + k} + \cdots+ \frac{a_n}{j + k} + \cdots = \lim_{j \to \infty}\frac{a_1/j}{1 + k/j} + \cdots+ \frac{a_n/j}{1 + k/j} +\cdots $$

Can someone please help me finish? I don't know if this a right way. Any help/hint/suggestion will be really appreciate it. Thank you in advance.

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The series $\sum a_k/(j+k)$ converges uniformly for all $j \in \mathbb{N}$ since $0 \leqslant a_k/(j+k) < a_k/k$ for k large and $\sum a_k/k$ converges.

For any $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that

$$\left|\sum_{k=1}^{\infty}\frac{a_k}{j+k}\right| \leqslant \left|\sum_{k=N+1}^{\infty}\frac{a_k}{j+k}\right|+\left|\sum_{k=1}^{N}\frac{a_k}{j+k}\right| \leqslant \frac{\epsilon}{2}+ \left|\sum_{k=1}^{N}\frac{a_k}{j+k}\right|.$$

With $N$ fixed, there exists $M \in \mathbb{N}$ such that if $j \geqslant M$ then for $k = 1,2,\ldots,N$

$$\frac{|a_k|}{j+k} < \frac{\epsilon}{2N},$$

and

$$\left|\sum_{k=1}^{N}\frac{a_k}{j+k}\right| \leqslant \sum_{k=1}^{N}\frac{|a_k|}{j+k} < \frac{\epsilon}{2}. $$

Hence, if $j \geqslant M$ then

$$\left|\sum_{k=1}^{\infty}\frac{a_k}{j+k}\right| <\epsilon.$$

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Assume that $a_k\geq 0$ for any $k> N$. Since $$ \lim_{j\to +\infty}\frac{a_m}{m+j} = 0, $$ it follows that: $$ \lim_{j\to +\infty}\left(\frac{a_1}{j+1}+\frac{a_2}{j+2}+\ldots+\frac{a_N}{j+N}\right) = 0,$$ so the first $N$ terms of the sequence do not really matter. Since $\sum_{k\geq 1}\frac{a_k}{k}$ is converging, we have: $$ \lim_{M\to +\infty}\sum_{k\geq M}\frac{a_k}{k}=0.$$ For any $j\geq N^{2\alpha}$, consider that: $$ \sum_{k>N}\frac{a_k}{k+j} = \sum_{k=N+1}^{N^{\alpha}}\frac{a_k}{k+j}+\sum_{k>N^{\alpha}}\frac{a_k}{k+j}\leq\frac{1}{N^{\alpha}}\sum_{k=N+1}^{N^{\alpha}}\frac{a_k}{k}+\sum_{k>N^{\alpha}}\frac{a_k}{k}, \tag{1}$$ and notice now that, as long as $\alpha\to +\infty$, both terms on the RHS of $(1)$ goes to zero.

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You can prove this by applying the Dominated Convergence Theorem. In the special case of sums of series, this theorem states:

Suppose that $\sum_{k=1}^\infty c_k$ converges, and for all $j,k$ we have $|A_{jk}| \le c_k$. Suppose, furthermore, that for all $k$ we have $A_{jk} \to b_k$ as $j \to \infty$. Then $$\sum_{k=1}^\infty A_{jk} \to \sum_{k=1}^\infty b_k$$ as $j \to \infty$.

In the particular problem from the question, use $A_{jk} := \frac{a_k}{j+k}$, $b_k := 0$, $c_k := \frac{|a_k|}{k}$.

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