Question

\[ \lim_{n\to\infty}\left(\frac{n^2+1}{\sqrt{n^6+1}}+\cdots+\frac{n^2+n}{\sqrt{n^6+n}}\right) = \underline{} \] rounded off to one decimal place.

Hint:

When each term of a sum behaves like \(1/n\) and there are \(n\) terms, the limiting value is usually finite and often equal to \(1\).

Answer 1

Correct Answer: 1

Step 1: Write the general term.
\[ \frac{n^2+k}{\sqrt{n^6+k}}, \quad k=1,2,\ldots,n \]

Step 2: Factor powers of \(n\).
\[ \frac{n^2+k}{\sqrt{n^6+k}} = \frac{n^2\left(1+\frac{k}{n^2}\right)}{n^3\sqrt{1+\frac{k}{n^6}}} \]

Step 3: Simplify the term.
\[ \frac{n^2+k}{\sqrt{n^6+k}} = \frac{1}{n}\cdot \frac{1+\frac{k}{n^2}}{\sqrt{1+\frac{k}{n^6}}} \]

Step 4: Find limiting behavior.
Since \(1\leq k\leq n\), we have
\[ \frac{k}{n^2}\to 0, \qquad \frac{k}{n^6}\to 0 \]

Step 5: Each term behaves like \(1/n\).
\[ \frac{n^2+k}{\sqrt{n^6+k}} \sim \frac{1}{n} \]

Step 6: Sum over \(n\) terms.
\[ \sum_{k=1}^{n}\frac{n^2+k}{\sqrt{n^6+k}} \sim \sum_{k=1}^{n}\frac{1}{n} = 1 \]

Step 7: Final answer.
\[ \boxed{1.0} \]