Sum of series question: $S_n = 1 + 1/4 + 1/9 + 1/16 + 1/25 + … + 1/n^2 < 2$

Question

Prove that for any nonzero natural $n$ it is true that $$S_n = 1 + 1/4 + 1/9 + 1/16 + 1/25 + … + 1/n^2 < 2.$$

I'm sort of at a loss here. I'm not sure if there exists some formula or method to sum this kind of series, since there is a variable ratio…

Please do not post in the imperative. If you have a question, please ask. Also, what have you tried? — JavaMan, May 05 '11 at 04:34
Moreover, your statement is clearly false: for example, if $n=1$ then your inequality does not hold. — Mariano Suárez-Álvarez, May 05 '11 at 04:36
People are answering a question that is not the one asked... (or, the challenge posed, rather) — Mariano Suárez-Álvarez, May 05 '11 at 04:39
Well if 1 + 1/4 + ..... + 1/n^2 < 2 - 1/n^1 then 1+1/4 + ..... + 1/n^2 + 1/(n+1)^2 < 2 - 1/n^2 + 1/(n+1)^2 = 2 +(-(n+1)^2/n^2(n+1)^2 + n^2/(n+1)^2)= 2 +(-2n - 1)/(n+1)^2n^2 < 2 - 1/(n+1)^2$. So it follows by induction. — fleablood, Oct 04 '16 at 00:15

score 12 · Answer 1 · answered May 05 '11 at 04:39

12

For $n>1$, a formula that will help is $$\frac{1}{n^2}<\frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n}.$$ This gives a telescoping series as an upper bound.

answered May 05 '11 at 04:39

Jonas Meyer

53,602

So I should just take the limit of (1/(n-1) - 1/n) as n approaches infinity? – user10504 May 05 '11 at 04:58
No, you should see what happens if you use this upper bound. For example, $1+\frac{1}{4}<1+1-\frac{1}{2}$. $1+\frac{1}{4}+\frac{1}{9}<1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}$. Simplify, continue the pattern and see what develops. You may also want to search for references to telescoping series. – Jonas Meyer May 05 '11 at 05:03

score 5 · Answer 2 · answered May 05 '11 at 13:44

5

you can compare with $1+\int_1^{\infty}n^{-2}$ (draw a picture) which is exactly $2$. then estimate any little bit of the error to get below $2$

answered May 05 '11 at 13:44

yoyo

9,943

score 4 · Answer 3 · answered May 05 '11 at 04:42

Hint: If you replace the $\frac{1}{n^2}$ terms after the first with the greater $\frac{1}{n(n-1)}$, you can use partial fractions and telescope the series. Alternately, there are difficult proofs that your series sums to $\frac{\pi^2}{6}\approx 1.64493 \lt 2$

davidlowryduda · Answer 4 · 2011-05-05T04:51:01.023

2

Hint: Consider the bound $\dfrac{1}{n^2} < \dfrac{1}{n-1} - \dfrac{1}{n}$.

edited May 05 '11 at 04:51

answered May 05 '11 at 04:36

davidlowryduda

91,687

2

Unfortunately, 2^n>(n+1)^2 for n large enough, so this does not bound the series. – Ross Millikan May 05 '11 at 04:40
Oh my gosh - you're right! Instead, consider the classic telescopic series! I'll edit my post. – davidlowryduda May 05 '11 at 04:42
@user: I should also note that you should check here for methods of actually calculating the series. – davidlowryduda May 05 '11 at 04:44

Sum of series question: $S_n = 1 + 1/4 + 1/9 + 1/16 + 1/25 + … + 1/n^2 < 2$

4 Answers4