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 a general term.
The $k$ th term is $\dfrac{n^2+k}{\sqrt{n^6+k}}$ for $k=1,2,\dots,n$.

Step 2: Pull out powers of $n$.
\[ \frac{n^2+k}{\sqrt{n^6+k}}=\frac{1}{n}\cdot\frac{1+\frac{k}{n^2}}{\sqrt{1+\frac{k}{n^6}}} \]

Step 3: See what the fraction does.
Since $1\le k\le n$, both $\frac{k}{n^2}$ and $\frac{k}{n^6}$ go to $0$. So the big fraction tends to $1$.

Step 4: Squeeze each term.
So every term is close to $\frac1n$ for large $n$.

Step 5: Add the $n$ terms.
\[ \sum_{k=1}^{n}\frac{n^2+k}{\sqrt{n^6+k}}\approx n\cdot\frac1n=1 \]

Step 6: Final value.
The limit is $1$.
\[ \boxed{1.0} \]