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Let $\sum_{k=1}^\infty a_k$ be a convergent series with nonnegative terms and let $a_{n_k}$ be a subsequence of {$a_k$}. Prove that the series $\sum_{k=1}^\infty a_{n_k}$ converges.

I am having trouble figuring out how to prove this and am pretty lost. My very first idea was to try and prove that it is increasing but don't know where I would take it from there. Any help would be appreciated!

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    The subseries is less than the series itself, so it converges by the comparison test. Am I missing something? – 2017-02-15
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    @TheCount To properly use the comparison test, you would need to first insert 0s into the series for the skipped elements. Which requires proof. And, depending on how the course is going, the comparison test might not have been proven yet. – 2017-02-15
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    @btilly That's why I didn't post as an answer. – 2017-02-15

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An alternate approach is to prove that $0 \leq \sum_{k=i}^j a_{n_k} \leq \sum_{k=n_i}^{n_j} a_k$. Now go back to the definition of the limit and take advantage of the fact that $i \leq n_i$.

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    This was what my comment was getting at. Nice formalisation. – 2017-02-15
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HINT Remember that the limit of the series is defined as the limit of the sequence of partial sums. In the case where the series only has non-negative terms, the sequence of partial sums is monotonically increasing. If you take away terms, you can only decrease the value of the sum and it's still monotonic. Now remember something about monotone and bounded sequences....

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Your idea is good. Let $S_l = \sum_{k=1}^l a_l$ be the sequence of partial sums of the original series and let $T_l = \sum_{k=1}^l a_{n_k}$ be the sequence of partial sums of your series. You know that $\sum_{k=1}^{\infty} a_k$ converges so let us denote the limit by $S = \sum_{k=1}^{\infty} a_k =\lim_{l \to \infty} S_l$. Since $a_l \geq 0$ both $S_l$ and $T_l$ are increasing and we have

$$ T_l = \sum_{k=1}^l a_{n_k} \leq \sum_{k=1}^{n_l} a_k =S_{n_l} \leq S $$

so $(T_l)_{l=1}^{\infty}$ is both increasing and bounded, hence tends to a limit and so $\sum_{k=1}^{\infty} a_{n_k}$ converges.

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    I am getting confused by your use of l in this – 2017-02-15
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    @JxxYsde3: I had a typo which I corrected. I use $l$ as an index in the partial sums sequence of the series. – 2017-02-15
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    I am confused by your use of l in $S_l$ – 2017-02-15
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Start from the definition of "converges" which says that there exists an $L$ such that for $\epsilon > 0$ there is an $n_0(\epsilon)$ such that for any $m>n_0$, $$ \left| \sum_{k=1}^m a_k - L \right| < \epsilon $$ Since the terms in the series are non-negative, $L$ is certainly an upper bound for all the partial $\sum_{k=1}^m a_k$.

Now consider for some ordered set $K$ of natural numbers and the series $a_{k_i}$. For all $m$, $$\sum_{i=1}^m a_{k_i} \leq \sum_{k=1}^{k_m}a_k \leq L$$ because the RHS is the LHS plus a (possibly empty) set of terms missing from the LHS, and the sum of those non-negative terms is non-negative. So the sequence $$ s_m = \sum_{i=1}^m a_{k_i} $$ is bounded from above (by $L$) and non-decreasing. Now consider the least upper bound $U$. (which must exist and be between $a_{k_1}$ and $L$. We can show that $U$ satisfies the definition of the limit of the sequence $s_m$ as follows:

Suppose there were some $\epsilon>$ such that there is no $n_0(\epsilon)$ satisfying the condition that for all $m>n_0$, $$ \left| s_m - U \right| = U-s_m < \epsilon $$ Then $U-\frac{\epsilon}{2}$ must be an upper bound of the sequence $s_m$. But we had taken $U$ to be the least upper bound; so there is a contradiction. Therefore, the definition of a limit must be satisfied for this $U$.