Let $\{a_n\}_{n\ge 1} $ be a sequence of real numbers satisfying $a_1 \ge 1$ and $a_{n+1}\ge a_n+1$ for all $n\ge 1.$ Then which of the following is necessarily true?

  1. the Series $\sum_{n=1}^\infty \frac{1}{a_n^2}$ diverges

  2. the sequence $\{a_n\}_{n\ge 1}$ is bounded

  3. the Series $\sum_{n=1}^\infty \frac{1}{a_n^2}$ converges

  4. the Series $\sum_{n=1}^\infty \frac{1}{a_n}$ converges

My attempt: $a_1\ge 1 \Rightarrow \frac{1}{a_1} \le 1 \Rightarrow \frac{1}{a_1^2} \le1 $

Given that $a_{n+1}\ge a_n+1$ for all $n\ge 1.\\\Rightarrow a_2 \ge a_1+1 \\\Rightarrow a_2 \ge 1+1=2 \Rightarrow\frac{1}{a_2^2}\le \frac{1}{2^2} $

similarly $a_3 \ge a_2+1 \\\Rightarrow a_3 \ge 2+1=3 \Rightarrow\frac{1}{a_3^2}\le \frac{1}{3^2} $

and so on

Let $S_n=\frac{1}{a_1^2}+\frac{1}{a_2^2}+\frac{1}{a_3^2}+........+\frac{1}{a_n^2} \\S_n \le 1+\frac{1}{2^2}+\frac{1}{3^2}+........+\frac{1}{n^2} \\ S_n \le \sum \frac{1}{n^2}$

Since by $p$ series test $\sum \frac{1}{n^2}$ converges, $S_n$ also converges

Therefore $\sum_{n=1}^\infty \frac{1}{a_n^2}$ converges

am I right?

  • $\begingroup$ Looks fine to me. $\endgroup$ Commented Mar 8, 2018 at 16:39
  • $\begingroup$ Yes, you are right. $\endgroup$ Commented Mar 8, 2018 at 16:39
  • $\begingroup$ Seems right. Some additional info- this series converges to $\dfrac{\pi ^2}{6}$ $\endgroup$ Commented May 9, 2019 at 15:47

1 Answer 1


Yes, you proved correctly that 3. holds. And, since there are sequences that satisfy the given condition, 1. doesn't hold. Also, 2. and 4. don't hold; take $a_n=n$ for each $n$.


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