Question

Find the limit of given series: $\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{1^{99}+2^{99}+\dots +n^{99}}{n^{100}}}=$

a)${\displaystyle \frac{1}{100}}$
b)$3$
c)${\displaystyle \frac{1}{3}}$
d)$1$

Answer 1

The sum of first n natural numbers to power one can be written as,
     $1^1+2^1+\dots +n^1=\frac{n(n+1)}{2}=\frac{n^2}{2}+\frac{n}{2}$
Similarly, the sum of first n natural numbers to power two can be written as,
     $1^2+2^2+\dots +n^2=\frac{n(n+1)(2n+1)}{6}$
     $1^2+2^2+\dots +n^2=\frac{(n^2+n)(2n+1)}{6}$
     $1^2+2^2+\dots +n^2=\frac{2n^3+n^2+2n^2+n}{6}$
     $1^2+2^2+\dots +n^2=\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}$
The sum of first n natural numbers to power three can be written as,
     $1^3+2^3+\dots +n^3={\left(\frac{n(n+1)}{2}\right)}^2$
     $1^3+2^3+\dots +n^3=\frac{n^2(n^2+2n+1)}{4}$
     $1^3+2^3+\dots +n^3=\frac{n^4+2n^3+n^2}{4}$
     $1^3+2^3+\dots +n^3=\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4}$
Generalizing this, we get
     $1^x+2^x+\dots +n^x=\frac{n^{x+1}}{x+1}+k_1{n}^x+k_2{n}^{x-1}+\dots$
Now substituting (x=100), we get
     $1^{99}+2^{99}+\dots +n^{99}=\frac{n^{99+1}}{99+1}+k_1{n}^{99}+k_2{n}^{99-1}+\dots$
Now the given expression becomes,
     $\underset{n\to \mathrm{\infty }}{lim}\frac{1^{99}+2^{99}+\dots +n^{99}}{n^{100}}=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{n^{99+1}}{99+1}+k_1{n}^{99}+k_2{n}^{99-1}+\dots }{n^{100}}$
     $$\Rightarrow \underset{n\to \mathrm{\infty }}{lim}\frac{1^{99}+2^{99}+\dots +n^{99}}{n^{100}}=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{n^{100}}{100}+k_1{n}^{99}+k_2{n}^{98}+\dots }{n^{100}}$$
     $$\Rightarrow \underset{n\to \mathrm{\infty }}{lim}\frac{1^{99}+2^{99}+\dots +n^{99}}{n^{100}}=\underset{n\to \mathrm{\infty }}{lim}\frac{n^{100}}{100n^{100}}+\frac{k_1{n}^{99}}{n^{100}}+\frac{k_2{n}^{98}}{n^{100}}+\dots$$
     $$\Rightarrow \underset{n\to \mathrm{\infty }}{lim}\frac{1^{99}+2^{99}+\dots +n^{99}}{n^{100}}=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{100}+\frac{k_1}{n}+\frac{k_2}{n^2}+\dots$$
Applying the limits, we get
     $\Rightarrow \underset{n\to \mathrm{\infty }}{lim}\frac{1^{99}+2^{99}+\dots +n^{99}}{n^{100}}=\frac{1}{100}+\frac{k_1}{\mathrm{\infty }}+\frac{k_2}{\mathrm{\infty }}+\dots$
We know, $\frac{1}{\mathrm{\infty }}\approx 0$, so
     $\Rightarrow \underset{n\to \mathrm{\infty }}{lim}\frac{1^{99}+2^{99}+\dots +n^{99}}{n^{100}}=\frac{1}{100}+0+0+\dots$
     $$\Rightarrow \underset{n\to \mathrm{\infty }}{lim}\frac{1^{99}+2^{99}+\dots +n^{99}}{n^{100}}=\frac{1}{100}$$
Hence, the correct option for the given question is option (a).
Answer - Option (a)
Note - In this type of question first we have to find the summation of a given series after that check the indeterminate form of limit, then substitute the limiting value you have in your answer.