Let $a_n={\frac1{\sqrt{n^2+1}}}+{\frac1{\sqrt{n^2+2}}}_{....}{\frac1{\sqrt{n^2+n+1}}} $

Find $\lim_{n\to\infty}a_n$.

I know the limit is zero. I also know that I need to use the squeeze theorem to solve it.

Now using ${\frac n{\sqrt{n^2+1}}}\le a_n \le {\frac n{\sqrt{n^2+n+1}}}$ as my upper and lower bounds doesn't yield 0, it yields 1.

It doesn't seem right but can I just use ${\frac 1{\sqrt{n^2+1}}}\le a_n \le {\frac 1{\sqrt{n^2+n+1}}}$ to solve ?


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    $\begingroup$ I don't understand the sequence: how does it "go" from $\;\frac1{\sqrt{x^2+1}}\;$ until $\;\frac1{\sqrt{x^2+x+1}}\;$ ?? $\endgroup$ – DonAntonio Nov 10 '13 at 16:23
  • $\begingroup$ Why do you think that the limit is zero? $\endgroup$ – njguliyev Nov 10 '13 at 16:25
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    $\begingroup$ I like how your right-handed side of the definition of $a_n$ does not contain n... Nicely defined sequence! $\endgroup$ – sve Nov 10 '13 at 16:34
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    $\begingroup$ So many comments...and I still cannot understand how the sequence is defined! $\endgroup$ – DonAntonio Nov 10 '13 at 16:41
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    $\begingroup$ Oh, dear! Because $\;\sqrt{n^2+n+1}\ge\sqrt{n^2+1}\;$ ...?! $\endgroup$ – DonAntonio Nov 10 '13 at 16:48

The limit is not 0, in fact you are almost there!

${\frac {n+1}{\sqrt{n^2+n+1}}}\le a_n \le {\frac {n+1}{\sqrt{n^2+1}}}$

For LHS, $\lim_{n\to\infty} \frac{n+1}{\sqrt{n^2+n+1}}=\lim_{n\to\infty}\frac{1+\frac{1}{n}}{\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}}=1$.

For RHS, $\lim_{n\to\infty} \frac{n+1}{\sqrt{n^2+1}}=\lim_{n\to\infty}\frac{1+\frac{1}{n}}{\sqrt{1+\frac{1}{n^2}}}=1$

Hence, by Squeeze Theorem, $\lim_{n\to\infty}a_n=1$

  • $\begingroup$ But every element is zero as x approach infinity so the sum of all the zeros should be zero... $\endgroup$ – GinKin Nov 10 '13 at 16:32
  • $\begingroup$ That's the wrong concept, it's like you are having infinitely many pieces of bits! $\endgroup$ – freak_warrior Nov 10 '13 at 16:33
  • $\begingroup$ This is a mess, from the original post: the first inequality (with $\;a_n\;$) is completely wrong, of course. $\endgroup$ – DonAntonio Nov 10 '13 at 16:45
  • $\begingroup$ @DonAntonio edited again! should be ok $\endgroup$ – freak_warrior Nov 10 '13 at 16:46
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    $\begingroup$ Well, that in fact is true, @freak_warrior...yet we must get used to be way more careful with these things of basic algebra. $\endgroup$ – DonAntonio Nov 10 '13 at 16:53

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