سوال ۳۱ کنکور ارشد علوم کامپیوتر ۱۳۹۸

Let

$S_n = \sum_{k=1}^{n}\frac{1}{\sqrt{n^{2}+k}} = \frac{1}{\sqrt{n^{2}+1}}+\frac{1}{\sqrt{n^{2}+2}}+\cdots+\frac{1}{\sqrt{n^{2}+n}}.$

The sum has exactly $n$ terms, and each index $k$ ranges over $1,2,\dots,n$.

Bound each term

For every $k$ with $1\le k\le n$ we have $n^{2}+1 \le n^{2}+k \le n^{2}+n$, so taking reciprocals of the square roots reverses the inequalities:

$\frac{1}{\sqrt{n^{2}+n}} \;\le\; \frac{1}{\sqrt{n^{2}+k}} \;\le\; \frac{1}{\sqrt{n^{2}+1}}.$

Sum the bounds (squeeze)

Adding these $n$ inequalities gives

$\frac{n}{\sqrt{n^{2}+n}} \;\le\; S_n \;\le\; \frac{n}{\sqrt{n^{2}+1}}.$

Now take limits of the two outer expressions by factoring $n$ out of each square root:

$\frac{n}{\sqrt{n^{2}+n}} = \frac{n}{n\sqrt{1+\tfrac{1}{n}}} = \frac{1}{\sqrt{1+\tfrac{1}{n}}} \;\xrightarrow[n\to\infty]{}\; 1,$

$\frac{n}{\sqrt{n^{2}+1}} = \frac{1}{\sqrt{1+\tfrac{1}{n^{2}}}} \;\xrightarrow[n\to\infty]{}\; 1.$

Both bounds tend to $1$, so by the squeeze theorem

$\lim_{n\to\infty} S_n = 1.$

This is option 2.

Exam trick: When a sum has $n$ terms each $\approx \tfrac{1}{n}$, squeeze the whole sum between $\tfrac{n}{\sqrt{n^2+n}}$ and $\tfrac{n}{\sqrt{n^2+1}}$ instead of trying to set up a Riemann integral. The factors of $n$ inside the radical vanish in the limit and you get $1$ instantly.