Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Find the limit of given 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)$1$

Last updated date: 14th Jul 2024
Total views: 60.9k
Views today: 1.60k
Verified
60.9k+ views
The sum of first n natural numbers to power one can be written as,
${{1}^{1}}+{{2}^{1}}+\ldots +{{n}^{1}}=\frac{n\left( n+1 \right)}{2}=\frac{{{n}^{2}}}{2}+\frac{n}{2}$
Similarly, the sum of first n natural numbers to power two can be written as,
${{1}^{2}}+{{2}^{2}}+\ldots +{{n}^{2}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$
${{1}^{2}}+{{2}^{2}}+\ldots +{{n}^{2}}=\frac{\left( {{n}^{2}}+n \right)\left( 2n+1 \right)}{6}$
${{1}^{2}}+{{2}^{2}}+\ldots +{{n}^{2}}=\frac{2{{n}^{3}}+{{n}^{2}}+2{{n}^{2}}+n}{6}$
${{1}^{2}}+{{2}^{2}}+\ldots +{{n}^{2}}=\frac{{{n}^{3}}}{3}+\frac{{{n}^{2}}}{2}+\frac{n}{6}$
The sum of first n natural numbers to power three can be written as,
${{1}^{3}}+{{2}^{3}}+\ldots +{{n}^{3}}={{\left( \frac{n\left( n+1 \right)}{2} \right)}^{2}}$
${{1}^{3}}+{{2}^{3}}+\ldots +{{n}^{3}}=\frac{{{n}^{2}}\left( {{n}^{2}}+2n+1 \right)}{4}$
${{1}^{3}}+{{2}^{3}}+\ldots +{{n}^{3}}=\frac{{{n}^{4}}+2{{n}^{3}}+{{n}^{2}}}{4}$
${{1}^{3}}+{{2}^{3}}+\ldots +{{n}^{3}}=\frac{{{n}^{4}}}{4}+\frac{{{n}^{3}}}{2}+\frac{{{n}^{2}}}{4}$
Generalizing this, we get
${{1}^{x}}+{{2}^{x}}+\ldots +{{n}^{x}}=\frac{{{n}^{x+1}}}{x+1}+{{k}_{1}}{{n}^{x}}+{{k}_{2}}{{n}^{x-1}}+\ldots$
Now substituting (x=100), we get
${{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}=\frac{{{n}^{99+1}}}{99+1}+{{k}_{1}}{{n}^{99}}+{{k}_{2}}{{n}^{99-1}}+\ldots$
Now the given expression becomes,
$\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{\frac{{{n}^{99+1}}}{99+1}+{{k}_{1}}{{n}^{99}}+{{k}_{2}}{{n}^{99-1}}+\ldots }{{{n}^{100}}}$
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{\frac{{{n}^{100}}}{100}+{{k}_{1}}{{n}^{99}}+{{k}_{2}}{{n}^{98}}+\ldots }{{{n}^{100}}}$
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{n}^{100}}}{100{{n}^{100}}}+\frac{{{k}_{1}}{{n}^{99}}}{{{n}^{100}}}+\frac{{{k}_{2}}{{n}^{98}}}{{{n}^{100}}}+\ldots$
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{100}+\frac{{{k}_{1}}}{n}+\frac{{{k}_{2}}}{{{n}^{2}}}+\ldots$
Applying the limits, we get
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\frac{1}{100}+\frac{{{k}_{1}}}{\infty }+\frac{{{k}_{2}}}{\infty }+\ldots$
We know, $\frac{1}{\infty }\approx 0$, so
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\frac{1}{100}+0+0+\ldots$
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\frac{1}{100}$
Hence, the correct option for the given question is option (a).