# If a series $\sum\lambda_n$ of positive terms is convergent, does the sequence $n\lambda_n$ converge to $0$? [closed]

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?

## closed as off-topic by YuiTo Cheng, Jendrik Stelzner, Paul Frost, José Carlos Santos, Xander HendersonMay 22 at 21:41

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• Of it didn't, then $\lambda_n$ would be on the order of $\frac{1}{n}$ and therefore the sum would diverge – Chickenmancer Jul 31 '17 at 16:42
• It is rather distressing to me how many smart people have gotten this innocuous problem completely wrong. – Steven Stadnicki Jul 31 '17 at 17:13
• @StevenStadnicki As a university student, I can say that I've seen very intelligent professors sometimes make "trivial" mistakes. We're all human, I guess. – MathematicsStudent1122 Jul 31 '17 at 20:27

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 $\sum_n\lambda_n<\infty$.

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.

• $\lambda_n > 0$? – 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
• @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