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I have a sequence $(a_n)_{n=1}^\infty$ with the following properties:

Monotonically decreasing

$a_n \geq 0$

$\sum_{n=1}^\infty a_n < \infty$

I have to show

$$\lim_{n \rightarrow \infty} n \ a_n = 0$$

My thoughts:

By theorem 3.23 in baby Rudin, if $\sum_{n=1}^\infty a_n < \infty$ then $\lim_{n \rightarrow \infty} a_n = 0$

Therefore, if I can show that $\sum_{n=1}^\infty n a_n < \infty$ then $\lim_{n \rightarrow \infty} n a_n = 0$

By theorem 3.24 in baby Rudin, A series of nonnegative terms converges if and only if its partial sums form a bounded sequence. So, I need to prove the sequence $n \ a_n$ has only nonnegative terms and the partial sum $\sum_{k=1}^n k a_k$ is bounded.

Now, the partial sum of n is bounded. By theorem 3.14 in baby Rudin, If the sequence $a_n$ is monotonic then the sequence converges iff it is bounded. I conclude $a_n$ is bounded. So if the sequence $a_n$ is bounded, it partial sum will be bounded.

Next, the product of two bounded series is bounded

$$\sum_{k=1}^n k \ \sum_{k=1}^n a_k = \sum_{k=1}^n k a_k \leq M$$

QED?

Any ideas?

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  • $\begingroup$ Showing that $\sum_{n=1}^\infty na_n < \infty$ is a bit strong. For example, the sequence $a_n = \frac1{n^2}$ satisfies your conditions, but $\sum_{n=1}^\infty na_n = \infty$ $\endgroup$ – Mathmo123 Oct 26 '14 at 18:22
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$$na_n=2\sum_{k=n/2}^na_n\leqslant2\sum_{k=n/2}^na_k\leqslant2\sum_{k=n/2}^\infty a_k\to0$$

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Let $\displaystyle S_N = \sum_{n=1}^N a_n$. Since $a_n \ge 0$, we have $$\frac{S_N}N \le \frac1N\sum_{n=1}^\infty a_n\to0\ \text{ as }N\to \infty$$ But $$\begin{align} S_N &\ge\sum_{n=1}^Na_N \quad\text{since $a_n$ is monotonically decreasing}\\&=Na_N\end{align}$$Therefore, $$0\le a_N \le \frac{S_N}N\to0$$

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  • We have $a_{1} > a_{2} > a_{3} > \cdots >a_{n} > a_{n+1} > \cdots$

  • $S_{n} = \sum_{k=1}^{n} a_{k}$. Then we have $\displaystyle\lim_{n \to \infty} a_{n}=0$ since $\sum_{n}a_{n}<\infty$

  • $S_{2n}=a_{1} + a_{2} + \cdots + a_{2n} = S_{n}+ a_{n+1} + a_{n+2} + \cdots +a_{2n} > S_{n}+n\cdot a_{2n}$

  • $0 < n \cdot a_{2n} < S_{2n}-S_{n}$ gives $\displaystyle \lim_{n \to \infty} n\cdot a_{2n}=0 \implies\lim_{n \to \infty} 2n \cdot a_{2n}=0$.

  • Similarly one can show $\displaystyle\lim_{n\to\infty} (2n+1)\cdot a_{2n+1}=0$

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