Prove that $a_{n+1}=a_{n} + 1/(n+1)^2$ is bounded from above

Question

Prove that $a_{n+1}=a_{n} + 1/(n+1)^2$ is bounded from above

What I have tried is $$a_n=1+1/4+\dots+1/n^2\leq 1+1+\dots +1=n$$

So I conclude that $a_n$ is bounded above by $n$. Does this sufficient to say that the sequence is bounded from above? Or it can only be said when the sequence is bounded from above by a constant but not any function of $n$?

Answer 1

hint: $a_{n+1} - a_n = \dfrac{1}{(n+1)^2} < \dfrac{1}{n}- \dfrac{1}{n+1}$. Can you see the teloscoping trick?

Answer 2

No, you proved your sequence is bounded by another sequence, you have to prove there is a real number which your sequence never exceeds.

The series in your example is sometimes called the p-series with $p=2$

Answer 3

I assume $a_0:=0$. As noted in the comments, your upper bound is not a constant. To get a constant, just put more terms in the sum. Precisely, take the upper bound to be $$ \sum_{i=1}^{\infty}\frac{1}{i^2}=\frac{\pi^2}{6}. $$ Note: You don't really need to know the value of this series, all you need to know is that it converges to some number $\alpha$ and take that $\alpha$.

Answer 4

Hints

• In general terms $ a_n = a_1 + \frac 1 3^2 + \frac 1 4^2 + ... + \frac {1}{ (n + 1)^2 }$
• The series $ \sum \frac{1}{n^2} $ converges.

Answer 5

We have $a_{n+1} - a_{n}=\frac{1}{(n+1)^2}$, so this is P-series (in this case $p=2$, your problem turns out to Basel problem