Question

Let $m$ and $k$ be positive integers$\left( {mk} \right)$.
Then
$\mathop {\lim }\limits_{n \to \infty } n\left\{ {{{\left( {1 + \dfrac{1}{n}} \right)}^m} + {{\left( {1 + \dfrac{2}{n}} \right)}^m} + ... + {{\left( {1 + \dfrac{k}{n}} \right)}^m} - k} \right\}$is:
1.$km$
2.$\dfrac{{k\left( {k + 1} \right)}}{2}m$
3.0
4.$\dfrac{{m\left( {m + 1} \right)k}}{2}$

Answer 1

Hint: Here, we are required to find the limit of a given sequence. We would try breaking the sequence as small as possible and instead of taking limits as a whole, we would find the limit of each bracket. Applying L’Hopital’s rule would help us to know the general limit of this sequence, and hence, we could find the sum of this sequence using progressions.

Complete step-by-step answer:
We are given two positive integers, $m$and $k$
And, we have to find:
$\mathop {\lim }\limits_{n \to \infty } n\left\{ {{{\left( {1 + \dfrac{1}{n}} \right)}^m} + {{\left( {1 + \dfrac{2}{n}} \right)}^m} + ... + {{\left( {1 + \dfrac{k}{n}} \right)}^m} - k} \right\}$
Now, let $\dfrac{1}{n} = y$
Hence, if $n \to \infty $ , then, $\dfrac{1}{n} = y \to {0^ + }$
$\mathop {\lim }\limits_{y \to {0^ + }} \dfrac{1}{y}\left\{ {{{\left( {1 + y} \right)}^m} + {{\left( {1 + 2y} \right)}^m} + ... + {{\left( {1 + ky} \right)}^m} - k} \right\}$
Now, if we let $k = 1$ and subtract it from all the brackets instead of as a whole, we get,
$\mathop {\lim }\limits_{y \to {0^ + }} \dfrac{1}{y}\left\{ {\left[ {{{\left( {1 + y} \right)}^m} - 1} \right] + \left[ {{{\left( {1 + 2y} \right)}^m} - 1} \right] + ... + \left[ {{{\left( {1 + ky} \right)}^m} - 1} \right]} \right\}$
Now, take $\dfrac{1}{y}$inside every bracket,
$ = \mathop {\lim }\limits_{y \to {0^ + }} \left\{ {\left[ {\dfrac{{{{\left( {1 + y} \right)}^m} - 1}}{y}} \right] + \left[ {\dfrac{{{{\left( {1 + 2y} \right)}^m} - 1}}{y}} \right] + ... + \left[ {\dfrac{{{{\left( {1 + ky} \right)}^m} - 1}}{y}} \right]} \right\}$
Hence, this can be written as:
L$ = \mathop {\lim }\limits_{y \to {0^ + }} \left[ {\dfrac{{{{\left( {1 + y} \right)}^m} - 1}}{y}} \right] + \mathop {\lim }\limits_{y \to {0^ + }} \left[ {\dfrac{{{{\left( {1 + 2y} \right)}^m} - 1}}{y}} \right] + ... + \mathop {\lim }\limits_{y \to {0^ + }} \left[ {\dfrac{{{{\left( {1 + ky} \right)}^m} - 1}}{y}} \right]$
Now, we would apply the L’Hopital’s rule, in which if we have an indeterminate form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$, we would differentiate the numerator and the denominator and then take the limit.
Now, applying L’Hopital’s rule,
We get,
$\mathop {\lim }\limits_{y \to {0^ + }} \left[ {\dfrac{{{{\left( {1 + ky} \right)}^m} - 1}}{y}} \right] = mk$
Hence, L can be written as:
L$ = 1 \times m + 2 \times m + ... + k \times m$
Here, this is in Arithmetic progression (A.P.)
Where, first term,$a = m$
Common difference,$d = 1$
Last term, $l = km$
Total number of terms,$n = k$
Hence, sum of given terms in an AP is:
$\dfrac{n}{2}\left( {a + l} \right)$
$ = \dfrac{k}{2}\left( {m + mk} \right)$
Taking $m$common, we get,
L$ = \dfrac{{k\left( {k + 1} \right)}}{2}m$
Hence,
$\mathop {\lim }\limits_{n \to \infty } n\left\{ {{{\left( {1 + \dfrac{1}{n}} \right)}^m} + {{\left( {1 + \dfrac{2}{n}} \right)}^m} + ... + {{\left( {1 + \dfrac{k}{n}} \right)}^m} - k} \right\} = \dfrac{{k\left( {k + 1} \right)}}{2}m$
Therefore, option (2) is the correct answer.

Note: We should know how to solve limits, apply L’Hopital’s rule and find the sum of $n$terms of an AP, to solve this question. Without the basic conceptual knowledge of limits, it is difficult to answer this question and also, we wouldn’t be able to reach our final answer without using Arithmetic progressions. Hence, application of the properties and formulas play a vital role to answer this question.