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…

$\endgroup$
  • $\begingroup$ Please do not post in the imperative. If you have a question, please ask. Also, what have you tried? $\endgroup$ – JavaMan May 5 '11 at 4:34
  • 1
    $\begingroup$ You don't seem to have a series... $\endgroup$ – Mariano Suárez-Álvarez May 5 '11 at 4:35
  • 1
    $\begingroup$ Moreover, your statement is clearly false: for example, if $n=1$ then your inequality does not hold. $\endgroup$ – Mariano Suárez-Álvarez May 5 '11 at 4:36
  • $\begingroup$ People are answering a question that is not the one asked... (or, the challenge posed, rather) $\endgroup$ – Mariano Suárez-Álvarez May 5 '11 at 4:39
  • $\begingroup$ @Mariano. You are right. I have deleted my posts. $\endgroup$ – JavaMan May 5 '11 at 4:40
10
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ So I should just take the limit of (1/(n-1) - 1/n) as n approaches infinity? $\endgroup$ – user10504 May 5 '11 at 4:58
  • $\begingroup$ 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. $\endgroup$ – Jonas Meyer May 5 '11 at 5:03
5
$\begingroup$

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 $2$

$\endgroup$
4
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ Unfortunately, 2^n>(n+1)^2 for n large enough, so this does not bound the series. $\endgroup$ – Ross Millikan May 5 '11 at 4:40
  • $\begingroup$ Oh my gosh - you're right! Instead, consider the classic telescopic series! I'll edit my post. $\endgroup$ – davidlowryduda May 5 '11 at 4:42
  • $\begingroup$ @user: I should also note that you should check here for methods of actually calculating the series. $\endgroup$ – davidlowryduda May 5 '11 at 4:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.