About Proving Convergence of Sequence

Question

Let ${b_n}$ be a sequence of real numbers defined by $b_1=1,b_2=0$, and $b_{n+2}=b_{n+1}+{b_n\over n^2}$. Prove that the sequence converges.

My solution: ${b_1}=1,b_2=0,b_3=1,b_4=1,b_5={10\over 9},...$

The sequence is increasing for $n\ge 4$, and I claim that the sequence is bounded above by 2.

Proof by induction: assume that $b_3<2,b_4<2,...b_{n-1}<2$, then $b_n-b_4=(b_n-b_{n-1})+(b_{n-1}-b_{n-2})+...+(b_5-b_4)={b_{n-2}\over (n-2)^2}+{b_{n-3}\over (n-3)^2}+...+{b_3\over 3^2} <2({1\over (n-2)^2}+{1\over (n-3)^2}+...+{1\over 3^2}) <2({\pi^2\over 6}-1-{1\over 4})<2({1\over 2})=1$

Hence $b_n<2$. By the monotone convergence theorem the sequence converges.

My question here is that the Basel identity that I used was not mentioned in my analysis course, so I wonder if anyone here could offer an alternative approach? Very much thanks!

Does this answer your question? Need to prove the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges — Anne Bauval, May 18 '23 at 17:00
Or this: Proving the sum of the reciprocals squared converges — Anne Bauval, May 18 '23 at 17:08
Knowing that ${1\over 1^2}+{1\over 2^2}+...+{1\over n^2}$ converges is not enough for my approach. I have to show that ${\pi^2\over 6}-1-{1\over 4}<{1\over 2}$ — Runyang Wang, May 18 '23 at 17:10
You don't have to show that. Just apply the same telescoping method as in these two duplicates (both found using Approach$0$): $\sum_{k=3}^{n-2}\frac1{k^2}<\frac12.$ — Anne Bauval, May 18 '23 at 17:18

score 1 · Answer 1 · answered May 19 '23 at 07:55

Let $S=\sum_{n=1}^{\infty}1/n^2.$ In order to obtain $2(S-1-\frac 1 4)<1,$ it is not necessary to know that $S=\frac {\pi^2}{6}.$ It suffices that $S<\frac 7 4 .$ We have $$S=1+\frac 1 4 +\sum_{n=3}^{\infty} \frac {1}{n^2}\;<\;1+\frac 1 4 +\sum_{n=3}^{\infty} \frac {1}{n-1} -\frac 1 n =$$ $$=1+\frac 1 4 +(\frac 1 2 -\frac 1 3)+ (\frac 1 3 -\frac 1 4)+(\frac 1 4 -\frac 1 5)+ ...=$$ $$=1+\frac 1 4 +\frac 1 2 =\frac 7 4.$$

About Proving Convergence of Sequence

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