Question

Let \(S_n=\displaystyle\sum_{k=1}^{n}\frac{n}{n^2+k}\), for \(n\in N\). Then the sequence \(\{S_n\}\) is ____.

• A. Convergent
• B. Divergent to \(\infty\)
• C. Bounded but not convergent
• D. Neither bounded nor diverges to \(\infty\)

Hint:

If a sequence has a finite limit, then it is convergent. Here, \(S_n\to 1\), so the sequence is convergent.

Answer 1

The Correct Option is A

Concept:
We are given: \[ S_n=\sum_{k=1}^{n}\frac{n}{n^2+k} \] To decide whether the sequence \(\{S_n\}\) is convergent or not, we study the limit of \(S_n\) as: \[ n\to\infty \] If the limit exists and is finite, then the sequence is convergent.

Step 1: Rewrite the general term.
Consider the term: \[ \frac{n}{n^2+k} \] Divide numerator and denominator by \(n^2\): \[ \frac{n}{n^2+k} = \frac{\frac{n}{n^2}}{\frac{n^2+k}{n^2}} \] \[ = \frac{\frac{1}{n}}{1+\frac{k}{n^2}} \] Therefore: \[ \frac{n}{n^2+k} = \frac{1}{n}\cdot \frac{1}{1+\frac{k}{n^2}} \]

Step 2: Rewrite the whole sum.
So, \[ S_n=\sum_{k=1}^{n}\frac{1}{n}\cdot \frac{1}{1+\frac{k}{n^2}} \] \[ S_n=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n^2}} \]

Step 3: Estimate each term.
Since: \[ 1\leq k\leq n \] we have: \[ \frac{1}{n^2}\leq \frac{k}{n^2}\leq \frac{n}{n^2} \] \[ \frac{1}{n^2}\leq \frac{k}{n^2}\leq \frac{1}{n} \] As \(n\to\infty\), \[ \frac{k}{n^2}\to 0 \] So, \[ 1+\frac{k}{n^2}\to 1 \] Hence, \[ \frac{1}{1+\frac{k}{n^2}}\to 1 \]

Step 4: Use bounding method.
Since \(k\geq 1\), \[ n^2+k\geq n^2+1 \] and since \(k\leq n\), \[ n^2+k\leq n^2+n \] Therefore: \[ \frac{n}{n^2+n}\leq \frac{n}{n^2+k}\leq \frac{n}{n^2+1} \] Now summing from \(k=1\) to \(n\): \[ \sum_{k=1}^{n}\frac{n}{n^2+n} \leq S_n \leq \sum_{k=1}^{n}\frac{n}{n^2+1} \] \[ n\cdot \frac{n}{n^2+n} \leq S_n \leq n\cdot \frac{n}{n^2+1} \] \[ \frac{n^2}{n^2+n} \leq S_n \leq \frac{n^2}{n^2+1} \]

Step 5: Take limits of both sides.
Now, \[ \lim_{n\to\infty}\frac{n^2}{n^2+n} = \lim_{n\to\infty}\frac{1}{1+\frac{1}{n}} =1 \] Also, \[ \lim_{n\to\infty}\frac{n^2}{n^2+1} = \lim_{n\to\infty}\frac{1}{1+\frac{1}{n^2}} =1 \] So, by squeeze theorem: \[ \lim_{n\to\infty}S_n=1 \]

Step 6: Final conclusion.
Since: \[ \lim_{n\to\infty}S_n=1 \] the sequence \(\{S_n\}\) has a finite limit. Therefore, the sequence is: \[ \text{Convergent} \] \[ \therefore \text{Correct Answer is (A)} \]