0
$\begingroup$

I have limit below which I try to find as written below. However professor told me I could not use the division by highest power because there infinitely many numbers in this sequence and I should use the squeeze theoreme instead? Why is it so? Where did I do the mistake?

\begin{align*}\lim_{n\to\infty}\log(1^1 +2^2 + ... + n^n) &= \lim_{n\to\infty}\log((n^n)(\dfrac{1^1}{n^n} +\dfrac{2^2}{n^n} + ... + 1))\\ &=\lim_{n\to\infty}\log(n^n)+\log(\dfrac{1^1}{n^n} +\dfrac{2^2}{n^n} + ... + 1)\\ &=\lim_{n\to\infty}n\log(n) + \lim_{n\to\infty} \log(\dfrac{1^1}{n^n}+\dfrac{2^2}{n^n} + ... + 1)\\ &= \lim_{n\to\infty} n\log(n) + 0 \end{align*}

$\endgroup$
  • 1
    $\begingroup$ You will probably want to rename $x$ for $n$. $\endgroup$ – user526015 Feb 8 at 14:37
  • $\begingroup$ @James Thanks, I did! $\endgroup$ – cris14 Feb 8 at 14:39
  • 1
    $\begingroup$ With the same argument you could show that $1 = (\frac 1n + \ldots + \frac 1n) \to 0$. $\endgroup$ – Martin R Feb 8 at 14:42
  • $\begingroup$ If you re-write the thing in the parentheses as $$\sum^n_{k=1} \frac{k^k}{n^n}$$ it should be fairly obvious that you can't say $$\lim_{n\to\infty} \left( \sum^{n}_{k=1} \frac{k^k}{n^n} \right) = \sum^n_{k=1} \left(\lim_{n\to\infty} \frac{k^k}{n^n}\right)$$ $\endgroup$ – User8128 Feb 8 at 14:43
  • 1
    $\begingroup$ @cris14 Are you satisfied with at least one of the answers? If so, please consider accepting one of them. If not, let us know what is still missing for you and how answers can be improved. $\endgroup$ – user526015 Feb 11 at 20:28
1
$\begingroup$

Let $(a_n),(b_n)$ be sequences. Then, $\lim a_n+b_n=\lim a_n+\lim b_n$ only holds if you know in advance that $\lim a_n$ and $\lim b_n$ exist.

This is not the case in what you are doing. Thus the separate calculation of the limits causes the mistake.

Of course in your case the sequences are the partial sums. In other words, if you are given infinite series $\sum a_n,\sum b_n$, then, $\sum a_n+b_n=\sum a_n+\sum b_n$ only holds if you know in advance that $\sum a_n$ and $\sum b_n$ exist (i.e. have a finite value).

$\endgroup$
  • 1
    $\begingroup$ This is one problem, but OP also uses $$\lim_{n\to\infty} \sum^n_{k=1} a_{k,n} \neq \sum^n_{k=1} \lim_{n\to\infty} a_{n,k}$$ which is not necessarily true. $\endgroup$ – User8128 Feb 8 at 14:42
  • $\begingroup$ @User8128 You are right, thanks! $\endgroup$ – user526015 Feb 8 at 14:42
0
$\begingroup$

Of course, you take the limit $n \to \infty$ and not $x \to \infty$. We have $$ \frac{1 + 2^2 + 3^3 + \cdots + n^n}{n^n} \le \frac{n^{n+1}}{n^n} \le n, $$ so that your computation yields $$ \lim_{n \to \infty} \frac{\ln(1 + 2^2 + \cdots + n^n)}{n \ln (n)} = 1, $$ or, more precisely, $$ \ln(1 + 2^2 + \cdots + n^n) = n \ln (n) + O(\ln(n)). $$

$\endgroup$

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.