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I need to find the limit of the sequence

$\dfrac{n + n^2 + n^3 + n^4 + \ldots + n^n}{1^n + 2^n + 3^n + 4^n + \ldots +n^n}$,

My strategy is to use Stolz's Cesaro theorem for this sequence.

Now, the numerator is given by :

$x_r = n^1+ n^2 +n^3 + \ldots +n^r$, so $x_{n+1} - x_{n} = n^{n+1}$

Similarly for denominator $y_r = 1^n + 2^n + 3^n +\ldots +r^n$, so $y_{n+1}- y_{n} = (n+1)^n$

Using Stolz Cesaro, this limit is equivalent to

$\displaystyle \lim \dfrac{n^{n+1}}{(n + 1)^n}$, which diverges to $ +\infty$,

However ans given to me is $\dfrac{e-1}{e}$, Can anyone tell where is the error in my solution ?

Thanks.

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  • $\begingroup$ The $\dfrac{n^{n+1}}{(n + 1)^n}$ does not converge, which is a condition for use of Cesaro-Stolz. There is a condition you can add so that it works the other way, see math.stackexchange.com/questions/2166597/… (of course in this case the condition will not be satisfied). $\endgroup$
    – Sil
    May 28, 2020 at 9:50
  • $\begingroup$ One can use this result. $\endgroup$ May 28, 2020 at 9:50

3 Answers 3

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Note that, as mentioned in the comments below, your computation of the ratio is incorrect. Regardless, the hypothesis of Stolz-Cesaro assumes that the limit $\lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n}$ exists. If it doesn't exist, it does not imply that the original limit does not exist.

A better way to approach is to write it as follows: $$ \frac{n + n^2 + \cdots + n^n}{1^n + 2^n + \cdots +n^n} = \frac{n^{-(n-1)} + n^{-(n-2)} + \cdots + n^{-1} + 1}{\left(\frac{1}{n}\right)^n + \left(\frac{2}{n}\right)^n + \cdots + \left(\frac{n-1}{n}\right)^n + 1} $$ As $n \to \infty$, clearly the numerator $\to 1$. For the denominator, see this.

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  • $\begingroup$ Thanks for the answer, just one doubt while evaluating the limit for $\dfrac{1^n + 2^n + \ldots +n^n}{n^n}$ why can't we use Stolz cesaro ? the conditons for the theorem are all satisfied so , limit should be $0$ $\endgroup$
    – user435638
    May 28, 2020 at 10:13
  • $\begingroup$ @sat091 We might be able to use Stolz-Cesaro, but your computation can't be right, as $\frac{1^n + 2^n + \cdots + n^n}{n^n} \geq \frac{n^n}{n^n} = 1$, so the limit can't be $0$. $\endgroup$ May 28, 2020 at 10:33
  • $\begingroup$ :Can you please try this with Stolz- Cesaro , your observation is correct the limit cannot be 0, but when I try using the theorem I get: $\dfrac{(n+1)^n}{(n+1)^{n+1} -n^n}$ and this converges to $0$ $\endgroup$
    – user435638
    May 28, 2020 at 10:38
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    $\begingroup$ @sat091 your computation of the ratio is incorrect. If $a_n = 1^n + 2^n + \cdots + n^n$, then $a_{n+1} = 1^{n+1} + 2^{n+1} + \cdots + n^{n+1} + (n+1)^{n+1}$. Thus, $a_{n+1} - a_n$ probably does not have a simple expression, and is definitely not $(n + 1)^n$, as: $$ a_{n+1} - a_n = (n+1)^{n+1} + \sum_{k=1}^n (k^{n+1} - k^n) \geq (n+1)^{n+1} > (n+1)^n $$ $\endgroup$ May 28, 2020 at 10:43
  • $\begingroup$ If this is incorrect then is my computation for ratio in the question is also wrong? because I used the same idea here . $\endgroup$
    – user435638
    May 28, 2020 at 11:04
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Divide nunberator and denominator by $n^n$. So your question consists of two limits, numerator and denominator, we'll deal with them separately.

For the numerator the limit would become $lim_{n \to \infty} 1+\frac{1}{n}+\ldots+\frac{1}{n^n} = 1*\frac{(1/n)^{n+1}-1}{(1/n)-1} = \lim_{h \to 0} \frac{h^{1+1/h} -1}{h-1} = -[e^{(1+1/h)\ln(h)} -1] = -[e^{-\infty}-1]=1$

For the denominator I'll prove a even more general limit, for any constant $k \neq 0$

$$\lim_{n to \infty}\frac{1^{kn}+2^{kn}+\ldots+n^{kn}}{n^{kn}}=\sum_{r=1}^{n} \frac{r^{kn}}{n^{kn}} = \sum_{r=0}^{n-1} \frac{(n-r)^{kn}}{n^{kn}}= \sum_{r=0}^{n-1} (1-r/n)^{kn} = \sum_{r=0}^{n-1} e^{-rk} = \frac{1}{1-e^{-k}} = \frac{e^k}{e^k-1}$$.

Here k=1, so final answer is $\frac{1}{(e/e-1)}=\frac{e-1}{e}$

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If you approximate the sums with the corresponding integrals it converges to 0 $$ L = \frac{\int_{1}^{n}n^x dx}{\int_{1}^{n} x^n dx} = \frac{n^{n+1} + n^n }{n^{n+1}\log n - \log n} = 0 $$

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