Knowing that sum from n=1 to infinity 1/(n^()alpha)
Knowing that $\sum_{n=1}^{\infty}\frac{1}{n^\alpha}$ converges for $\alpha>1$ and diverges for $\alpha\le1$, determine the convergence of: $$ \sum_{n=1}^{\infty} \frac{n\sin\frac{1}{n}}{\sqrt{n^2+1}}. $$
Explanation of Method
Both the numerator's $n\sin\frac1n$ and the denominator's $\sqrt{n^2+1}$ approach clean limits (respectively $1$ and $n$) as $n\to\infty$, but neither correction is strong enough to change the overall order: the term behaves exactly like $\frac1n$. Since $\frac1n$ sits precisely on the harmonic divergence boundary, the limit comparison test settles the matter.
- Approximate the denominator $\sqrt{n^2+1}\sim n$.
- Approximate the numerator $n\sin\frac1n\to1$.
- Apply the limit comparison test against $\sum1/n$.
Step 1 · Approximate the denominator
$$ \sqrt{n^2+1} = n\sqrt{1+\tfrac1{n^2}} \sim n \quad\text{for large } n. $$
Step 2 · Approximate the numerator
Using $\sin t = t - \frac{t^3}{6}+O(t^5)$ with $t=\frac1n$: $n\sin\frac1n = 1-\frac1{6n^2}+O\!\left(\frac1{n^4}\right) \to 1$.
Step 3 · Combine and apply the limit comparison test
So $a_n = \dfrac{n\sin\frac1n}{\sqrt{n^2+1}} \sim \dfrac1n$. With $b_n=\dfrac1n$:
$$ \lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\frac{n^2\sin\frac1n}{\sqrt{n^2+1}} = \lim_{n\to\infty}\frac{n\sin\frac1n}{\sqrt{1+\frac1{n^2}}} = \frac{1}{1} = 1. $$
Positive and finite; since $\sum\frac1n$ diverges, by the limit comparison test $\sum_{n=1}^{\infty}\dfrac{n\sin\frac1n}{\sqrt{n^2+1}}$ diverges as well.
Summary Table
| Quantity | Value |
|---|---|
| Denominator asymptotics | $\sqrt{n^2+1}\sim n$ |
| Numerator asymptotics | $n\sin\frac1n\to1$ |
| General term $a_n$ | $\sim\dfrac1n$ |
| Conclusion | Diverges |
Additional Notes
- A precise expansion gives $a_n = \frac1n - \frac2{3n^3}+O\!\left(\frac1{n^5}\right)$, confirming that only the leading $\frac1n$ term matters for convergence — the correction terms decay fast enough to be irrelevant.
- Both the numerator and denominator individually have "nice" limits ($n\sin\frac1n\to1$, $\sqrt{n^2+1}/n\to1$), which can create the illusion that the whole fraction settles to a finite nonzero limit rather than decaying — but the outer factor of $\frac1n$ (from dividing by $\sqrt{n^2+1}\sim n$) is what actually controls convergence.
- This is structurally similar to Problem 38: sine of a small quantity contributes a factor asymptotically equal to that quantity, and here it is exactly this substitution that reduces the whole expression to the harmonic rate.
Result
The series diverges: