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Let $(x_n)_{n \in \Bbb N}$ be a decreasing sequence such that its series converges, want to show that $\displaystyle \lim_{n \to \infty} n x_n = 0$.

Ok I don't even know where to start. I need a direction please! Thankyou!

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Are the $x_n$ positive real numbers? – Lord_Farin Oct 29 '12 at 11:49
i dont know if xn are positive real nos. – d13 Oct 29 '12 at 11:54
up vote 5 down vote accepted

The sequence $(x_n)$ is nonincreasing and the series $\sum\limits_nx_n$ converges, hence $x_n\geqslant0$ for every $n$. Otherwise, there exists $n$ such that $x_n\lt0$, and $x_k\leqslant x_n$ for every $k\geqslant n$, hence $$ \sum\limits_{k=n+1}^mx_k\leqslant(m-n)x_n\to-\infty $$ when $m\to\infty$, which is absurd. Furthermore, the series $\sum\limits_kx_k$ converges hence $\sum\limits_{k\gt n}x_k\to0$ when $n\to\infty$. But $$ \sum\limits_{k\gt n}x_k\geqslant\sum\limits_{k=n+1}^{2n}x_k\geqslant nx_{2n}\geqslant0, $$ hence $nx_{2n}\to0$ hence $nx_n\to0$.

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thankyou very much! i appreciate ur help. – d13 Oct 29 '12 at 14:39
i understood the last part pretty much well, but i still dont get the part which has to do with m. how did u get an inequality with m and why does m tends to infinity. – d13 Oct 29 '12 at 16:03
If $k\gt n$, $x_k\leqslant x_n$. Add this from $k=n+1$ to $k=m$, you get the inequality in the post. Next, this is valid for every $m\gt n$ and, if $x_n\lt0$, $(m-n)x_n\to-\infty$ when $m\to+\infty$, hence the rest $\sum\limits_{k\gt n}x_k$ would not go to zero when $n\to+\infty$, something which contradicts the summability of the series. – Did Oct 29 '12 at 17:48

Just another approach. Since $\{x_n\}_{n\in\mathbb{N}}$ id decreasing and its associated series converges, we have $x_n\geq 0$ for every $n\in\mathbb{N}$ (otherwise, $\lim_{n\to +\infty}x_n < 0$ and the series cannot converge). Assume now the existence of a positive real number $\alpha$ such that $$ n\, x_n \geq \alpha $$ for an infinite number of positive natural numbers $n$; let $A=\{n_1,n_2,\ldots\}$ be the set of such natural numbers. Let now $a_0=0,a_1=n_1$, $a_2$ be the minimum element of $A$ greater than $2a_1$, $a_3$ be the minimum element of $A$ greater than $2a_2$ and so on. We have: $$\sum_{n=1}^{+\infty}x_n \geq \sum_{k=1}^{+\infty}(a_k-a_{k-1})x_{a_k} \geq\sum_{k=1}^{+\infty}\frac{\alpha}{2}=+\infty,$$ that is clearly a contradiction, so $$\lim_{n\in\mathbb{N}} (n\,x_n) = 0$$ must hold.

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