3

Find the limit :

$$\lim_{ n \to \infty }\left( (\frac{1}{n})^{17}+(\frac{2}{n})^{17}+(\frac{3}{n})^{17}+...+(\frac{n}{n})^{17} \right) =?$$

My Try:

$$\lim_{ n \to \infty } (\frac{1}{n})^{17}=(\frac{2}{n})^{17}=(\frac{3}{n})^{17}=...=(\frac{k}{n})^{17}=0\\\lim_{ n \to \infty } (\frac{n}{n})^{17}=1$$

so : $$\lim_{ n \to \infty }\left( (\frac{1}{n})^{17}+(\frac{2}{n})^{17}+(\frac{3}{n})^{17}+...+(\frac{n}{n})^{17} \right) =1$$

is it right ?

Almot1960
  • 4,782
  • 16
  • 38
  • 1
    Your approach is not correct, When you sum together many things which tend to 0 their sum can be very large. For $n=5$ the sum is $1.02$, and at $n=400$ the sum is $22.7$ – Hugh Mar 02 '17 at 09:42

3 Answers3

9

Like The limit of a sum $\sum_{k=1}^n \frac{n}{n^2+k^2}$.

$$\lim_{n\to\infty}\dfrac1n\sum_{r=1}^n\left(\dfrac rn\right)^{17}=\int_0^1x^{17}dx$$

Community
  • 1
lab bhattacharjee
  • 274,582
0

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim_{n \to \infty}\sum_{k = 1}^{n}\pars{k \over n}^{17} & = \lim_{n \to \infty}\pars{{1 \over n^{17}}\sum_{k = 1}^{n}k^{17}} = \lim_{n \to \infty}{\pars{n + 1}^{17} \over \pars{n + 1}^{17} - n^{17}} = \bbx{\ds{\infty}} \end{align}

Note that the numerator is a $17$-degree polynomial while the Denominator is a $16$-degree one !!!.

Felix Marin
  • 89,464
0

Denote the sum as $s_n$, then given any $M > 0$, we have there exists large enough $m > 0$ such that, $$ s_k > M,\quad k \ge m $$

For example, $m = \Big{\lceil}2M/(1 - \frac{1}{2^{1/17}})\Big{\rceil}$ (let $m_1=\Big{\lceil}\frac{m}{2^\frac{1}{17}}\Big{\rceil}$, clearly $\big{(}\frac{m_1}{m}\big{)}^{17} \ge \frac{1}{2}$ and $m - m_1 > 2M$, hence $s_m > \frac{1}{2}\cdot(2M)=M$).

Therefore, $\lim_{n\rightarrow \infty}s_n=\infty$. (Definition 3.3 in http://math.mit.edu/~apm/ch03.pdf)

BTW, @labbhattacharjee, (Sorry I can not make a comment on your answer directly), I have some doubts about some arguments in the answer. I think we can only get

$$ \lim_{n \rightarrow \infty}f(n)g(n) = \lim_{n\rightarrow \infty}f(n)\lim_{n\rightarrow \infty}g(n) $$

when both $g(n)$ and $f(n)$ have finite limit.

For example,

$$ 1 = \lim_{n\rightarrow \infty}\frac{n}{n + 1} \neq \lim_{n\rightarrow \infty}n\lim_{n\rightarrow\infty}\frac{1}{n+1} = \infty\cdot 0 $$

Long Gong
  • 1