Squeeze Theorem for the Limit of an n-Term Sum

The squeeze (sandwich) theorem

If $a_n \le b_n \le c_n$ for all large $n$ and the outer sequences converge to the same value,

$\lim_{n\to\infty} a_n = \lim_{n\to\infty} c_n = L \quad\Longrightarrow\quad \lim_{n\to\infty} b_n = L,$

then the trapped middle sequence is forced to the common limit $L$. The whole art is choosing $a_n$ and $c_n$ that are (a) genuinely above/below $b_n$ and (b) easy to take the limit of.

The pattern: a sum with a growing number of terms

Many konkur limits look like a finite sum whose number of terms depends on $n$:

$S_n = \sum_{k=1}^{n} f(n,k), \qquad f(n,k) = \frac{1}{\sqrt{n^2+k}},\ \frac{n}{n^2+k},\ \frac{k}{n^2+k},\ \dots$

You cannot just send each term to $0$: there are $n$ terms, and "$n$ things each near $0$" can sum to anything. Direct summation is usually impossible too. The squeeze theorem is the standard escape, and it works whenever the terms are monotone in $k$ so the first and last term are the extremes.

How to apply it

1. Count the terms. Write $S_n=\sum_{k=1}^{n} f(n,k)$ and note there are exactly $n$ of them.
2. Find the extreme terms. For $1\le k\le n$, if $f(n,k)$ decreases in $k$ then $f(n,n)\le f(n,k)\le f(n,1)$; replace every term by the smallest (lower bound) and by the largest (upper bound).
3. Multiply by the count. Summing $n$ identical bounds gives

$n\cdot \min_k f \;\le\; S_n \;\le\; n\cdot \max_k f.$

4. Take the limit of both ends. If they agree, $S_n$ is squeezed to that common value.

Worked mini-example

Evaluate $\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\frac{n}{n^2+k}$.

For $1\le k\le n$ the denominator satisfies $n^2+1 \le n^2+k \le n^2+n$, so each of the $n$ terms is sandwiched:

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

Add the $n$ inequalities:

$\frac{n\cdot n}{n^2+n} \;\le\; S_n \;\le\; \frac{n\cdot n}{n^2+1}, \qquad\text{i.e.}\qquad \frac{n^2}{n^2+n} \;\le\; S_n \;\le\; \frac{n^2}{n^2+1}.$

Both outer fractions $\to 1$ (divide top and bottom by $n^2$), so $\displaystyle\lim_{n\to\infty} S_n = 1$.

The same three moves — bound, multiply by $n$, take the limit — solve $\sum 1/\sqrt{n^2+k}$, whose bounds $\tfrac{n}{\sqrt{n^2+n}}$ and $\tfrac{n}{\sqrt{n^2+1}}$ both tend to $1$.

Exam trick: the gap between the largest and smallest term comes from the part of the denominator that varies ($+k$, with $1\le k\le n$), which is negligible next to the dominant $n^2$. As soon as both bounds collapse to the same value, you are done — no antiderivative, no Riemann sum needed.

Pitfall: the bound must be the whole sum, so remember the factor of $n$ from adding $n$ copies — bounding only a single term (and forgetting to multiply by the term count) is the most common error and gives a limit of $0$ instead of the true value.