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Let $\lambda_n>0, n\in\mathbb{N}$, with $\sum_n \lambda_n<+\infty$.

Can I conclude that $n\lambda_n\to 0$?

In this question and this question and their answers, it is shown that this is true if $\lambda_n$ are decreasing. What happens if $\lambda_n$ are not decreasing?

robjohn
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Anton
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2 Answers2

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No.

Define $\lambda_n$ by stating that $\lambda_{2^n}=2^{-n}$ and $\lambda_k=2^{-k}$ for other values of $k$.

Then $2^n\lambda_{2^n}=1$ so there is no convergence to $0$.

It is evident however that $\sum_n\lambda_n<\infty$.

drhab
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Consider $$ \lambda_n=\left\{\begin{array}{} \frac1n&\text{if $n=k^2$ for some $k\in\mathbb{Z}$}\\ \frac1{n^2}&\text{if $n\ne k^2$ for any $k\in\mathbb{Z}$}\\ \end{array}\right. $$ Then, when $n=k^2$, $$ n\lambda_n=1 $$ yet $$ \sum_{n=1}^\infty\lambda_n=2\zeta(2)-\zeta(4) $$

However, if we have $\lambda_k\ge\lambda_{k+1}$, then $$ \lim_{n\to\infty}n\lambda_n=0 $$ Suppose not. Then there is an $\epsilon\gt0$ so that for any $n$, there is an $N\ge n$ so that $N\lambda_N\ge\epsilon$. Then, because of the monotonicity, we have $$ \begin{align} \sum_{k=N/2}^{N}\lambda_k &\ge\sum_{k=N/2}^{N}\frac\epsilon{N}\\ &\ge\frac\epsilon2 \end{align} $$ and since we can choose $n$ as large as we want, there is a limitless set of sequences of terms whose sum is at least $\frac\epsilon2$. That is, we can choose $n_{j+1}=2N_j+2$ so that $N_{j+1}/2\ge n_{j+1}/2\gt N_j$, so that the intervals $[N_j/2,N_j]$ are disjoint and $\sum\limits_{k=N_j/2}^{N_j}\lambda_k\ge\frac\epsilon2$. Therefore, $$ \sum_{k=1}^\infty\lambda_k=\infty $$ Note: this latter argument is similar to this answer.

robjohn
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    $\lambda_n > 0$? – Alex Ortiz Jul 31 '17 at 17:00
  • @AOrtiz: okay, if you must quibble about that, I have changed the other terms to be bigger. It still converges, and the limit is not $0$. We need monotonicity, or something similar, to guarantee that $n\lambda_n$ converges to $0$. – robjohn Jul 31 '17 at 17:12
  • Is the sequence $\lambda_n$ in the second part of your answer the same as the sequence in the first part? I am also confused by your claim. $\lambda_n\to 0$ because the series converges. You probably want to claim $\lim n\lambda_n = 0$? – Alex Ortiz Jul 31 '17 at 17:16
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    @AOrtiz: no. In the first part, I have given a sequence where $\lim\limits_{n\to\infty}n\lambda_n\ne0$ yet the sum converges. In the latter part, I have shown that the conclusion is true if the sequence is monotonic. – robjohn Jul 31 '17 at 17:19
  • I don't understand the downvote. Other than the fact that I missed that the terms must be $\gt0$ instead of $\ge0$, there was nothing wrong with my initial answer. I have even shown a positive result for monotonic sequences. – robjohn Jul 31 '17 at 17:44
  • Neither do I. The difference between $>0$ and $\geq0$ is actually not relevant here (+1). – drhab Jul 31 '17 at 18:08
  • Can you elaborate on what you mean "and since we can choose $n$ as large as we want, there is a limitless set of sequences of terms whose sum is at least $\epsilon/2$." I am having difficulty parsing this since it is not precise. – Alex Ortiz Jul 31 '17 at 18:39
  • @AOrtiz: I have added a bit of explanation to the answer. Please let me know if that doesn't answer your question. – robjohn Jul 31 '17 at 19:04
  • @AOrtiz: if you still have questions, you can also look at the answer I cited, which follows the same argument. – robjohn Jul 31 '17 at 22:41