Find $$\lim_{n \to \infty}\sum_{k=1}^{n}\left(1+\frac{2k}n\right)^3\frac 2n$$

I’m not really sure how to start. Does the ratio test apply?

EDIT: Sorry, horrible use of Latex.


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  • $\begingroup$ There is no $x$ in your question. $\endgroup$ – Thomas Andrews Aug 14 at 18:34
  • 6
    $\begingroup$ The above is the Reimann sum of $$\int_1^3 x^3\,dx=\frac{3^4}4-\frac{1^4}4=20.$$ $\endgroup$ – Thomas Andrews Aug 14 at 18:38
  • 1
    $\begingroup$ Thanks! I need to review. $\endgroup$ – Mr. Snrub Aug 14 at 18:56

I'll ignore Thomas Andrews' nice answer of a Riemann sum and use power sums in the form $s_a(n) =\sum_{k=1}^n k^a $ with

$s_0(n) = n,\\ s_1(n) = \frac12 n(n+1) = \frac12 n^2+\frac12 n,\\ s_2(n) = \frac16 n(n+1)(2n+1) =\frac13 n^3+\frac12 n^2+\frac16 n,\\ s_3(n) = \frac14 n^2(n+1)^2 = \frac14 n^4+\frac12 n^3+\frac14 n^2 $.

$\begin{array}\\ t(n) &=\sum_{k=1}^{n}\left(1+\dfrac{2k}n\right)^3\dfrac 2n\\ &=\dfrac 2n\sum_{k=1}^{n}\left(1+3\dfrac{2k}n+3\dfrac{(2k)^2}{n^2}+\dfrac{(2k)^3}{n^3}\right)\\ &=\dfrac 2n\left(\sum_{k=1}^{n}1+\dfrac{6}n\sum_{k=1}^{n}k+\dfrac{12}{n^2}\sum_{k=1}^{n}k^2+\dfrac{8}{n^3}\sum_{k=1}^{n}k^3\right)\\ &=\dfrac 2n\left(n+\dfrac{6}n(\frac12 n^2+\frac12 n)+\dfrac{12}{n^2}(\frac13 n^3+\frac12 n^2+\frac16 n)+\dfrac{8}{n^3}(\frac14 n^4+\frac12 n^3+\frac14 n^2)\right)\\ &=2+6+\frac{6}{n}+8+\frac{12}{n}+\frac{4}{n^2}+4+\frac{8}{n}+\frac{4}{n^2}\\ &=20+\frac{26}{n}+\frac{8}{n^2}\\ &\to 20 \qquad\text{as } n \to \infty\\ \end{array} $

The Riemann sum method is definitely easier.

Note that if we can use $\lim_{n \to \infty}\dfrac1{n^{a+1}}s_a(n) =\dfrac1{a+1} $, we get the Riemann sum answer immediately.

  • $\begingroup$ Nice (+1). This is a cool way to solve the given problem. $\endgroup$ – clathratus Aug 14 at 19:26

Avoiding the Riemann sum technique, we will use the known sums:

$$\begin{align}\sum_{k=1}^n k &= \frac{n(n+1)}{2}\\ \sum_{k=1}^n k^2 &= \frac{n(n+1)(2n+1)}{6}\\ \sum_{k=1}^n k^3 &= \frac{n^2(n+1)^2}{4} \end{align}$$

Now for fixed $n$ the value is:

$$\lim_{n \to \infty}\sum_{k=1}^{n}\left(1+\frac{2k}n\right)^3\frac 2n=\frac{2}{n^4}\sum_{k=1}^{n}(n+2k)^3$$

We can come up with a closed formula for $\sum_{k=1}^{n}(n+2k)^3,$ namely:

$$\begin{align}\sum_{k=1}^{n}(n+2k)^3 &= \sum_{k=1}^{n}\left(n^3+6n^2k + 12nk^2+8k^3\right)\\ &=n^4 + 6n^2\sum_{k=1}^{n} k +12n\sum_{k=1}^n k^2 + 8\sum_{k=1}^n k^3\\ &=n^4+6n^2\frac{n(n+1)}{2}+12n\frac{n(n+1)(2n+1)}{6}+8\frac{n^2(n+1)^2}{4}\\ &=n^2\left(n^2+3n(n+1)+2(n+1)(2n+1)+2(n+1)^2\right)\\ &=n^2(10n^2+13n+4) \end{align}$$

Multiplying by $\frac{2}{n^4}$ you get the value:


so you can easily get the limit of this as $n\to\infty.$

  • 1
    $\begingroup$ Essentially the same as mine, and done in a nicer way. $\endgroup$ – marty cohen Aug 14 at 19:32
  • $\begingroup$ Minor error: you didn't multiply the final 4 by 2 - should be $8/n^2$. Had to reconcile our differences. $\endgroup$ – marty cohen Aug 14 at 19:39
  • $\begingroup$ Thanks, fixed. @martycohen $\endgroup$ – Thomas Andrews Aug 14 at 20:56

Remember the Riemann Sum definition for the integral: $$\int_a^b f(x)dx=\lim_{n\to\infty}\frac{b-a}{n}\sum_{k=0}^{n}f\left(a+\frac{k}{n}(b-a)\right).$$ With this, we can set $b-a=2$ and $a=1$ so that $b=3$. We can also set $f(x)=x^3$ so that $$S=\lim_{n\to\infty}\frac2n\sum_{k=0}^n\left(1+\frac{2k}{n}\right)^3=\int_a^b f(x)dx=\int_1^3x^3dx=20$$


A definite integral $\int_a^b f(x)dx$ is a limit of Riemann sums defined by

$$ \int_a^b f(x) dx = \lim_{n\to\infty} \sum_{i=1}^n f(x_i) \Delta x$$

where the interval $[a,b]$ is divided into $n$ subintervals, and $x_i$ are the evaluation points.

You are given that

$$\lim_{n \to \infty}\sum_{k=1}^{n}\left(1+\frac{2k}n\right)^3\frac 2n$$


$$\Delta x = \frac{b-a}{n}=\frac{2}{n} \implies b-a=2$$


$$f(x_i)=\left(1+\frac{2k}n\right)^3 \implies x_i = \left(1+\frac{2k}n\right), ~ f(x)=x^3$$

since $x_i$ starts at $1$, we let $a=1$ and see that $b = a+2=3$. Therefore

$$\lim_{n \to \infty}\sum_{k=1}^{n}\left(1+\frac{2k}n\right)^3\frac 2n = \int_1^3 x^3 dx = 20$$


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