How to prove that if a series converges, then the sequence of it's terms multiplies by the inverses of their positions converges How to prove that for a descending sequence $(a_n)_{n \geq 1}$, if $\lim_{n \to \infty} \sum_1^n a_n$ converges to something finite, then $\lim_{n \to \infty} n a_n = 0$
A similar problem was asked here but it did not include that the sequence, in the case of the problem the sequence of the differences, is descending.
I attempted using the definition. If I take an $\epsilon$, I can try to prove that after a point $n a_n$ will always be smaller than $\epsilon$, because after a point there has to be an $n$ for which $n a_n < \epsilon$, because otherwise the series wouldn't converge as $\sum \frac{\epsilon}{n}$ doesn't converge, but unfortunately just because $a_n$ is deceasing, that doesn't mean $n a_n$ is also decreasing so I was stuck, because basically the product can go lower than $\epsilon$ and then go back over
 A: Hint: First of all argue that since $a_n$ is decreasing and the series $\sum_{k=1}^\infty a_n$ converges it must be the case that $a_n \geq 0$ for all $n\in \mathbb{N}$. Then you can apply Cauchy condensation test to get that the sequence $2^ka_{2^k}$ tends to $0$. Finally, use the fact that for $n$ between $2^{k}$ and $2^{k+1}$ we have
$$2^{k}a_{2^{k+1}} \leq n a_n \leq 2^{k+1}a_{2^{k}}$$
The squeeze theorem implies that $na_n\to 0$.
EDIT: Consider $s_n=a_1+\dots+a_n$. Since $s_n$ converges we know that it is a Cauchy sequence. Take $\varepsilon > 0$. There exists a $N\in \mathbb{N}$ such that for all $n,m\geq N$ we have $|s_n-s_m|<\varepsilon$. In particular, $|s_n - s_N| < \varepsilon$ for all $n\geq N$. Expanding the last inequality and using the fact that $a_n$ is decreasing we get
$$(n-N)|a_n| \leq |s_n - s_N| < \varepsilon$$
This shows that $na_n \to 0$.
A: Note that $a_n \geq 0$ because $\sum a_n$ converges.
Then : $(n-n/2)a_n \leq \sum_{k=n/2}^{n} a_k$ ($a_n$ is decreasing). The right end side converge to 0 because $\sum a_n$ converges.
A: One way to prove this would be to find two sequences $(u_n) \to 0$ and $(l_n) \to 0$, such that $u_n ≥ na_n ≥ l_n$ for all $n$, and then apply the squeeze theorem to conclude that $(na_n) \to 0$.
First, as a lemma, prove that a descending sequence whose partial sums converge to a finite limit must be non-negative.  (This is most easily proved via the contrapositive, i.e. by showing that the partial sums of a descending sequence that includes a negative element must diverge to negative infinity.)  This implies two useful facts:

*

*Since $(a_n)$ is non-negative, so is $(na_n)$.  Thus we can choose $l_n = 0$ as our lower bound sequence.


*Since $(a_n)$ is non-negative, the sequence of partial sums $s_n = \sum_{i=1}^n a_i$ must be monotone increasing.
Now, while $(s_n)$ is guaranteed (by the exercise statement) to converge to a limit, that limit is not guaranteed to be zero.  Thus, even though we do have $s_n = \sum_{i=1}^n a_i ≥ \sum_{i=1}^n a_n = na_n$ for all $n$, we unfortunately cannot simply choose $u_n = s_n$ and declare the exercise done.
However, maybe we can salvage this somehow?
Here's one way: First observe that, if $(s_n) \to c$, then the "half-stepped" sequence $(s_{\lfloor n/2 \rfloor})$ also converges to the same limit $c$, just slower.  (Note that $\lfloor n/2 \rfloor = 0$ when $n = 1$, but we can just define $s_0 = 0$ to handle that edge case.)
Since $(s_n)$ is monotone increasing, we also have $s_{\lfloor n/2 \rfloor} ≤ s_n ≤ c$ for all $n$, which in turn implies that $(s_n - s_{\lfloor n/2 \rfloor}) \to 0$.  (You can prove this using the squeeze theorem and the fact that $0 ≤ s_n - s_{\lfloor n/2 \rfloor} ≤ c - s_{\lfloor n/2 \rfloor}$.)  And since multiplying a convergent sequence elementwise by a constant factor just multiplies its limit by the same factor, we also have $(2(s_n - s_{\lfloor n/2 \rfloor})) \to 0$.
Now we can observe that $$2(s_n - s_{\lfloor n/2 \rfloor}) = 2\sum_{i=\lfloor n/2 \rfloor+1}^n a_i ≥ 2\sum_{i=\lfloor n/2 \rfloor+1}^n a_n = 2\lceil n/2 \rceil a_n ≥ na_n ≥ 0.$$
Thus we can apply the squeeze theorem, with upper bound $u_n = 2(s_n - s_{\lfloor n/2 \rfloor})$ and lower bound $l_n = 0$, to show that $(na_n) \to 0$. $\square$
(Ps. This trick would be much less cumbersome if we were dealing with continuous functions rather than discrete sequences, since then we wouldn't have to worry about rounding $n/2$ to integer values.)
