# If $\{a_n\}$ is nonincreasing and $a_n\ge1/n$ for infinitely many $n$s then $\sum a_n$ diverges

Let $\sum a_n$ be a series with infinitely many integers $n$ such that $a_n>1/n$. If $\{a_n\}$ is decreasing show that $\sum a_n$ is divergent.

My Attempt: In these kinds of problems I usually try to prove that $\lim a_n \not =0.$ The first hypothesis of the problem says that there is a subsequence $a_{n_k}>\frac{1}{n_k}.$ Now if we consider the sum $$\sum_{i=1}^{n_m}a_i\geq \sum_{k=1}^{m}a_{n_k}>\sum_{k=1}^{m}\frac{1}{n_k}.$$ I am not sure how to use the fact that $a_n$ is decreasing here to get a series on the right that diverges. Any hints/advice will be much appreciated.

• Well, without the condition that $\{a_n\}$ is decreasing it is false. The sequence $(0,1,0,\frac{1}{2},0,0,0,\frac{1}{4},0,\dots,0,\frac{1}{8},\dots)$ will have its series converge to $2$ despite having infinitely many $n$ such that $a_n>\frac{1}{n}$. (note the difference between "infinitely many $n$" and "all $n$") – JMoravitz Jan 19 '18 at 21:18
• @JMoravitz and there is no exclusion here of negative values for $a_n$ so $1-\frac 12+\frac 12-\frac 13+\frac 13 + \dots$ needs to be considered too. The decreasing condition is essential. If $a_n\gt 1$ for all $n$ then it is easy. So assume otherwise ... – Mark Bennet Jan 19 '18 at 21:25
• You cannot expect to show that $\lim a_n\ne0$. For instance, you could have $a_n=2/n$ for all $n$. – Andrés E. Caicedo Jan 19 '18 at 21:30
• What would be another approach? – model_checker Jan 19 '18 at 21:34
• The Integral test should work. If $a_{n_1},a_{n_2},\cdots$ denote the infinite subsequence such that $a_{n_i}>\frac 1{n_i}$ $\forall i$ then we note that, for $n_i>k>n_{i+1}$ we must have $a_k>\frac 1{n_{i+1}}$. So use the $a_{n_i}$ as endpoints for a Riemann sum approximation of the (divergent) integral of $f(x)=\frac 1x$. – lulu Jan 19 '18 at 21:35

Presume all the conditions are satisified, but suppose the opposite: $\sum a_n$ converges. Because of the convergence, starting from some $n_0$, we have: $\sum_{i=m}^n a_i\lt\frac{1}{2}$ for all $m,n\gt n_0$ (Cauchy criterion with $\varepsilon=\frac{1}{2}$).

Now find one $n\ge 2n_0$ such that $a_n\gt\frac{1}{n}$. Then, due to $a_n$ decreasing, we have $a_{n_0+1}\ge a_{n_0+2}\ge\cdots\ge a_n\gt\frac{1}{n}$, so $\sum_{i=n_0+1}^n a_i\gt\frac{n-n_0}{n}=1-\frac{n_0}{n}\ge\frac{1}{2}$ as we chose $n\ge 2n_0$ - a contradiction!

Hint:

If $a_n$ is decreasing and $\sum a_n$ converges then $\lim_{n\to \infty} na_n = 0$

In response to downvote

The decreasing condition forces $na_n \to 0$ if $\sum a_n$ converges. Hence, it would be impossible for there to exist a subsequence $a_{n_k} > 1 / n_k$. Therefore, $\sum a_n$ diverges when that subsequence exists.

By convergence and monotonicity we have $2n a_{2n} < 2\sum_{k=n+1}^{2n} a_k < \epsilon$ for sufficiently large $n$, implying $2n a_{2n} \to 0$ and with a similar argument for the subsequence $(2n+1) a_{2n+1}$ we can show that $na_n \to 0$.

• Hypotheses are $a_n$ decreasing and there exists subsequence for which $n_k a_{n_k} > 1$. Assume $\sum a_n$ converges then we get a contradiction in that $n a_n \to 0 \implies n_k a_{n_k} \to 0$ for any subsequence. Therefore $\sum a_n$ converges. – RRL Jan 19 '18 at 21:34
• the point is that OP want a proof of the hint. – Tsemo Aristide Jan 19 '18 at 21:35
• I disagree. This is a hint and the crux of the argument in proof by contradiction. The fact that $na_n \to 0$ is where the decreasing condition fits in. And I can give more of a hint in that direction idf necessary. – RRL Jan 19 '18 at 21:38
• $\sum a_n$ with $a_n$ decreasing implying $n a_n \to 0$ is well known. – RRL Jan 19 '18 at 21:40
• Correct. I just put a $2$ in front of the sum to fix since by Cauchy the sum can be made smaller than $\epsilon/2$ for sufficiently large $n$. Thanks. – RRL Jan 19 '18 at 22:18

The hypothesis is that there exists an increasing sequence $(m(k))$ of positive integers such that, for every $k$, $$a_{m(k)}\geqslant\frac1{m(k)}$$ The sequence $(a_n)$ is nonincreasing hence, for every $k$ and every $n$ between $m(k)+1$ and $m(k+1)$, $$a_n\geqslant a_{m(k+1)}\geqslant \frac1{m(k+1)}$$ This implies that, for every $k\leqslant\ell$, $$\sum_{n=m(k)+1}^{m(\ell+1)} a_n\geqslant\sum_{i=k}^\ell\frac{m(i+1)-m(i)}{m(i+1)}\geqslant\frac1{m(\ell+1)}\sum_{i=k}^\ell (m(i+1)-m(i))=\frac{m(\ell+1)-m(k)}{m(\ell+1)}$$ When $\ell\to\infty$, $m(\ell+1)\to\infty$ hence the RHS converges to $1$. This proves that, for every $k$, $$\sum_{n=m(k)+1}^\infty a_n\geqslant1$$ The rest of a converging series converges to zero hence the series $\sum a_n$ diverges.