3
$\begingroup$

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…

  • 0
    Please do not post in the imperative. If you have a question, please ask. Also, what have you tried?2011-05-05
  • 1
    You don't seem to have a series...2011-05-05
  • 1
    Moreover, your statement is clearly false: for example, if $n=1$ then your inequality does not hold.2011-05-05
  • 0
    People are answering a question that is not the one asked... (or, the challenge posed, rather)2011-05-05
  • 0
    @Mariano. You are right. I have deleted my posts.2011-05-05
  • 0
    See also [here](http://math.stackexchange.com/questions/174828/a-probably-trivial-induction-problem-sum-2nk-2-lt1) and [here](http://math.stackexchange.com/questions/1220203/proving-1-frac14-frac19-cdots-frac1n2-leq-2-frac1n-for).2016-10-03
  • 0
    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.2016-10-04

4 Answers 4

10

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.

  • 0
    So I should just take the limit of (1/(n-1) - 1/n) as n approaches infinity?2011-05-05
  • 0
    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.2011-05-05
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 $2$

4

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 $\frac{\pi^2}{6}\approx 1.64493 \lt 2$

2

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

  • 2
    Unfortunately, 2^n>(n+1)^2 for n large enough, so this does not bound the series.2011-05-05
  • 0
    Oh my gosh - you're right! Instead, consider the classic telescopic series! I'll edit my post.2011-05-05
  • 0
    @user: I should also note that you should check [here](http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-n-1-infty-frac1n2) for methods of actually calculating the series.2011-05-05