Question

Find the limit of given sum of a series $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=$
(a) $\dfrac{1}{100}$
(b) $3$
(c) $\dfrac{1}{3}$
(d) $1$

Answer 1

Hint: Here we find the sum of first natural numbers to power , then use some algebraic operation and also take the help of limit properties to solve the given problem.

Complete step-by-step solution -
The sum of first n natural numbers to power one can be written as,
${{1}^{1}}+{{2}^{1}}+\ldots +{{n}^{1}}=\dfrac{n\left( n+1 \right)}{2}=\dfrac{{{n}^{2}}}{2}+\dfrac{n}{2}$
Similarly, the sum of first n natural numbers to power two can be written as,
${{1}^{2}}+{{2}^{2}}+\ldots +{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$
${{1}^{2}}+{{2}^{2}}+\ldots +{{n}^{2}}=\dfrac{\left( {{n}^{2}}+n \right)\left( 2n+1 \right)}{6}$
${{1}^{2}}+{{2}^{2}}+\ldots +{{n}^{2}}=\dfrac{2{{n}^{3}}+{{n}^{2}}+2{{n}^{2}}+n}{6}$
${{1}^{2}}+{{2}^{2}}+\ldots +{{n}^{2}}=\dfrac{{{n}^{3}}}{3}+\dfrac{{{n}^{2}}}{2}+\dfrac{n}{6}$
The sum of first n natural numbers to power three can be written as,
${{1}^{3}}+{{2}^{3}}+\ldots +{{n}^{3}}={{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}$
${{1}^{3}}+{{2}^{3}}+\ldots +{{n}^{3}}=\dfrac{{{n}^{2}}\left( {{n}^{2}}+2n+1 \right)}{4}$
${{1}^{3}}+{{2}^{3}}+\ldots +{{n}^{3}}=\dfrac{{{n}^{4}}+2{{n}^{3}}+{{n}^{2}}}{4}$
${{1}^{3}}+{{2}^{3}}+\ldots +{{n}^{3}}=\dfrac{{{n}^{4}}}{4}+\dfrac{{{n}^{3}}}{2}+\dfrac{{{n}^{2}}}{4}$
Generalizing this, we get
${{1}^{x}}+{{2}^{x}}+\ldots +{{n}^{x}}=\dfrac{{{n}^{x+1}}}{x+1}+{{k}_{1}}{{n}^{x}}+{{k}_{2}}{{n}^{x-1}}+\ldots$
Now substituting ($x=99$), we get
${{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}=\dfrac{{{n}^{99+1}}}{99+1}+{{k}_{1}}{{n}^{99}}+{{k}_{2}}{{n}^{99-1}}+\ldots$
Now the given expression becomes,
$\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{\dfrac{{{n}^{99+1}}}{99+1}+{{k}_{1}}{{n}^{99}}+{{k}_{2}}{{n}^{99-1}}+\ldots }{{{n}^{100}}}$
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{\dfrac{{{n}^{100}}}{100}+{{k}_{1}}{{n}^{99}}+{{k}_{2}}{{n}^{98}}+\ldots }{{{n}^{100}}}$
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{n}^{100}}}{100{{n}^{100}}}+\dfrac{{{k}_{1}}{{n}^{99}}}{{{n}^{100}}}+\dfrac{{{k}_{2}}{{n}^{98}}}{{{n}^{100}}}+\ldots$
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{100}+\dfrac{{{k}_{1}}}{n}+\dfrac{{{k}_{2}}}{{{n}^{2}}}+\ldots$
Applying the limits, we get
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\dfrac{1}{100}+\dfrac{{{k}_{1}}}{\infty }+\dfrac{{{k}_{2}}}{\infty }+\ldots$
We know, $\dfrac{1}{\infty }$ tends to zero, so
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\dfrac{1}{100}+0+0+\ldots$
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\dfrac{1}{100}$
Hence, the correct option for the given question is option (a).