ISI Entrance test

Say, $L\,=\,\displaystyle \lim_{n \to \infty}\left[100n-n \sum_{k=1}^{100}f \left(1+\frac{k}{n}\right) \right]$

Notice that, $L\,=\,\displaystyle \lim_{n \to \infty}\left[100n-n \sum_{k=1}^{100}f \left(1+\frac{k}{n}\right) \right]\,=\,\lim_{n \to \infty} n\cdot\left[100-\sum_{k=1}^{100}f \left(1+\frac{k}{n}\right) \right]\,=\, \lim_{n \to \infty} \left[\dfrac{100- \sum_{k=1}^{100}f \left(1+\frac{k}{n}\right)}{\dfrac{1}{n}} \right]$ . Since, $f(1)=1$ ,therefore, $100\,=\,\sum_{k=1}^{100} 1\,=\,\sum_{k=1}^{100}f(1)$ .

So, $L\,=\, \displaystyle \lim_{n \to \infty} \left[\dfrac{\sum_{k=1}^{100}f(1)- \sum_{k=1}^{100}f \left(1+\frac{k}{n}\right)}{\dfrac{1}{n}} \right] \,=\, \lim_{n \to \infty} \sum_{k=1}^{100} \left[\dfrac{f(1) \, - \,f \left(1+\frac{k}{n}\right)}{\dfrac{1}{n}} \right] \,=\, \lim_{n \to \infty} \sum_{k=1}^{100}\,k\,\cdot \left[\dfrac{f(1) \, - \,f \left(1+\frac{k}{n}\right)}{\dfrac{k}{n}} \right]$

$=\, \displaystyle - \lim_{n \to \infty} \sum_{k=1}^{100}\,k\,\cdot \left[\dfrac{ f \left(1+\frac{k}{n}\right) \, - \, f(1)}{\dfrac{k}{n}} \right]$ .

Since, the summation variable $k$ has nothing to do with $n$ , therefore, it is legitimate to take limit inside summation. Doing so would yield : $L\,=\, - \displaystyle \sum_{k=1}^{100} \,k\,\cdot \lim_{n \to \infty} \left[\dfrac{ f \left(1+\frac{k}{n}\right) \, - \, f(1)}{\dfrac{k}{n}} \right]$ .

Now, notice that $k \, \in \, \{1,2,3,..,100\}$ , i.e $k$ is finite. Thus, as $\displaystyle n \to \infty$ , $\displaystyle \dfrac{k}{n} \to 0$ . Hence,

$L\,=\, - \displaystyle \sum_{k=1}^{100} \,k\,\cdot \lim_{t \to 0} \left[\dfrac{ f(1+t) \, - \, f(1)}{t} \right]$ , where $\dfrac{k}{n}\,=\,t$ .

Notice that, $\displaystyle \lim_{t \to 0} \left[\dfrac{ f(1+t) \, - \, f(1)}{t} \right]$ appears to be $f^{\prime} (1)$ . It can be ascertained that $f^{\prime} (1)$ does exist and is equal to $10$ .

So, $L\,=\, - \displaystyle \sum_{k=1}^{100} \,k\,\cdot \lim_{t \to 0} \left[\dfrac{ f(1+t) \, - \, f(1)}{t} \right]\,=\, - \displaystyle \sum_{k=1}^{100} \,k\,\cdot f^{\prime }(1)\,=\, - \sum_{k=1}^{100} \,k\,\cdot 10$ .

Therefore, $L\,=\, -10 \cdot \frac{(100)(100+1)}{2}\,=\,-50500$ .

Hence, $\frac{L}{10}\,=\,-5050$ .