limit without expansion

Question

I was solving this limit instead of using sum of expansion i did this and got zero but answer is 1/5 by expansion why this is wrong $$\lim_{n\rightarrow \infty } \frac{1^4 + 2^4 + \ldots + n^4}{n^5}$$

$$\lim_{n\rightarrow \infty } \frac{n^4( \frac{1^4}{n^4} + \frac{2^4}{n^4} +\ldots + 1 )}{n^5}$$

$$= \lim_{n\rightarrow \infty } \frac{( \frac{1^4}{n^4} + \frac{2^4}{n^4} +\ldots + 1 )}{n}$$

$$= \frac{(0 + 0 +0 \ldots + 1 )}{n} = 0$$ this is how i got  ,

Answer 1

You were going right till the very last step. You need to apply the limit after computing the sum. The limit converts to Riemann integral:

$$\lim_{n\to \infty}\sum_{k=1}^{n} \frac{k^4}{n^5} = \int_{0}^{1} x^4 dx = \frac{1}{5}$$

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Since you are not familiar with integration, another method is to find $\sum_{k=1}^{n} k^4$ explicitly. Only the leading coefficient, ie the coefficient of $n^5$ needs to be found.

$$\sum_{k=1}^nk^4=\frac1{30}n(n+1)(2n+1)(3n^2+3n-1)$$

So we can write $\sum_{k=1}^nk^4 = \frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} -\frac{n}{30}$. Thus if you compute the limit

$$\lim_{n\to \infty}\sum_{k=1}^n\frac{k^4}{n^5} = \lim_{n\to\infty} \frac{\frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} -\frac{n}{30}}{n^5}\\ \lim_{n\to\infty} \frac{\frac{1}{5} + \frac{1}{2n} + \frac{1}{3n^2} -\frac{1}{30n^4}}{1} = \frac{1}{5}$$

Answer 2

Also, you can use Stolz: https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem $$\lim_{n\rightarrow+\infty}\frac{1^4+2^4+...+n^4}{n^5}=\lim_{n\rightarrow+\infty}\frac{n^4}{n^5-(n-1)^5}=\frac{1}{5}.$$

Answer 3

If $m > 1$ then $k^m \lt \int_k^{k+1} x^m dx \lt (k+1)^m $.

Summing the left inequality from $1$ to $n-1$,

$\begin{array}\\ \sum_{k=1}^{n-1}k^m &\lt \sum_{k=1}^{n-1}\int_k^{k+1} x^m dx\\ &= \int_1^{n} x^m dx\\ &= \dfrac{x^{m+1}}{m+1}\big|_1^{n}\\ &< \dfrac{n^{m+1}-1}{m+1}+(n+1)^m\\ &< \dfrac{n^{m+1}}{m+1}\\ \text{so}\\ \sum_{k=1}^{n}k^m &< \dfrac{n^{m+1}}{m+1}+n^m\\ \text{so}\\ \dfrac1{n^{m+1}}\sum_{k=1}^{n}k^m &< \dfrac{1}{m+1}+\dfrac1{n}\\ \end{array} $

Similarly, summing the right inequality from $0$ to $n-1$,

$\begin{array}\\ \sum_{k=0}^{n-1}(k+1)^m &\gt \sum_{k=0}^{n-1}\int_k^{k+1} x^m dx\\ &= \int_0^{n} x^m dx\\ &= \dfrac{x^{m+1}}{m+1}\big|_0^{n}\\ &= \dfrac{n^{m+1}}{m+1}\\ \text{so}\\ \sum_{k=1}^{n}k^m &> \dfrac{n^{m+1}}{m+1}\\ \text{so}\\ \dfrac1{n^{m+1}}\sum_{k=1}^{n}k^m &> \dfrac{1}{m+1}\\ \end{array} $

Therefore $0 \lt \dfrac1{n^{m+1}}\sum_{k=1}^{n}k^m - \dfrac{1}{m+1} \lt \dfrac1{n} $ so $\lim_{n \to \infty} \dfrac1{n^{m+1}}\sum_{k=1}^{n}k^m =\dfrac1{m+1}$.

Note: Nothing original here, though I took pains to make the result come about as painlessly as possible.

Answer 4

You can’t do this step

$$\lim_{n\rightarrow \infty } \frac{ \frac{1^4}{n^4} + \frac{2^4}{n^4} +...+1 }{n}\color{red}{=\frac{0 + 0 +0 +...+1 }{n}= 0} $$

since you have infinitely many terms which tend to 0 and their sum not necessarily is $0$.

Think to $\sum_1^{\infty} \frac1k$ which diverges.

Answer 5

The problem is that while each term tends to zero, the number of terms tends to infinity at the same time.

The rule “the limit of a sum is the sum of the limits” only applies if you have a finite sum with a fixed number of terms (each of which has a finite limit).

(Also, you can't really write $$\lim_{n \to \infty} (\dots) = (\text{some expression involving $n$})$$ since there can be no $n$ left in your expression after you have let $n \to \infty$. The symbol $n$ stands for the same number everywhere, so you can't selectively let some $n$ tend to $\infty$ while leaving other $n$ unaffected.)

Answer 6

If we define $$ s(n):=\sum_{k=0}^nk^4 $$ you want to calculate $$\lim\limits_{n\to \infty}\frac{\sum\limits_{k=0}^nk^4}{n^5} $$ You say $$\lim\limits_{n\to \infty}\left(\sum_{k=0}^n \frac{ k^4}{n^5}\right)=\lim\limits_{n\to \infty}\left(\sum_{k=0}^n \lim\limits_{m\to \infty}\frac{ k^4}{m^5}\right) $$ but this transformation is generally not allowed.

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We know that $s$ is a polynomial of degree $5$. If you don't know let us assume that this is the case. So we define $$ p(n):=b_0+b_1n+b_2n(n-1)+b_3n(n-1)(n-2)+b_4n(n-1)(n-2)(n-3)+b_5n(n-1)(n-2)(n-3)(n-4) $$ and we can calculate the the $b_i$ successively:

$$ p(0)=s(0) \implies b_0=0\\ p(1)=s(1) \implies b_0+b_1=1 \implies b_1=1\\ p(2)=s(1)+2^4 \implies b_0+2b_1+2b_2=17\implies b_2=\frac{15}{2} $$ and further $$ b_3=\frac{25}{3}\\ b_4=\frac{5}{2}\\ b_5=\frac{1}{5} $$

So we have

$$s(n)=\frac{1}{5}n^5+p_4x^4+p_3x^3+p_2x^2+p_1x+p_0$$

The coefficients $p_4,\ldots,p_0$ are easily to calculate but we don't need there exact values.

So $$\lim\limits_{n\to \infty}\frac{\sum_{k=0}^nk^4}{n^5}\\ =\lim\limits_{n\to \infty}\frac{\frac{1}{5}n^5+p_4x^4+p_3x^3+p_2x^2+p_1x+p_0}{n^5}=\lim\limits_{n\to \infty}\left(\frac{1}{5}+p_4\frac{1}{n}+p_3\frac{1}{n^2}+p_2\frac{1}{n^3}+p_1\frac{1}{n^4}+p_0\frac{1}{n^5}\right)\\ =\lim\limits_{n\to \infty}1 +p_4\lim\limits_{n\to \infty}\frac{1}{n} +p_3\lim\limits_{n\to \infty}\frac{1}{n^2} +p_2\lim\limits_{n\to \infty}\frac{1}{n^3} +p_1\lim\limits_{n\to \infty}\frac{1}{n^4} +p_0\lim\limits_{n\to \infty}\frac{1}{n^5}\\ =1+0+0+0+0+0$$

Here we apply the well known laws about limits.

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If you don'believe that $s(n)$ is a polynomial of degree $5$ the you can proof it by induction (or look up the formula at Wikipedia).

You can calculate the missing coefficient $p_i$ and get the polynomial $${{n^5}\over{5}}+{{n^4}\over{2}}+{{n^3}\over{3}}-{{n}\over{30}}$$

Then you use

$$\left({{\left(n+1\right)^5}\over{5}}+{{\left(n+1\right)^4}\over{2}}+{{ \left(n+1\right)^3}\over{3}}-{{n+1}\over{30}}\right) - \left({{n^5}\over{5}}+{{n^4}\over{2}}+{{n^3}\over{3}}-{{n}\over{30}}\right)= (n+1)^4 $$

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(without proof)

Let's calculate the values $$s(0),\;s(1),\;s(2),\ldots$$
the differences $$s(1)-s(0),\;s(2)-s(1),\;s(3)-s(2),\;\ldots$$ of these values, the differences of these differences and so on, as shown in the table below.

 0     1    17    98   354   979  2275  4676  8772 15333 
    1    16    81   256   625  1296  2401  4096  6561
      15    65   175   369   671  1105  1695  2465
         50   110   194   302   434   590   770
            60    84   108   132   156   180
               24    24    24    24    24   
                   0     0     0     0

If $s(n)$ is a polynomial we eventually reach a line that only contains $0$. If the degree of $s$ is $d$ then it is the line $d+2$. Note that the left site of the triangle consists of the numbers

$$0, 1, 15, 50, 60, 24, 0,\ldots$$

Also note that

$$\frac{0}{0!}, \frac{1}{1!}, \frac{15}{2!}, \frac{50}{3!}, \frac{60}{4!}, \frac{24}{5!}, 0\ldots$$

are the coefficients $b_i$ of the Newton polynomial $p(x)$.

So the coefficient $b_5$ that we needed for the calculation of our limit can be found quickly by calculation these differences.

A proof can be found in

van der Waerden
Algebra 1
Achte Auflage der Modernen Algebra
Springer Verlag Berlin - Heidelberg - New York  1971
pp 88-92

or anywhere in the internet.