Based on the general term a_n, conclude that
Based on the general term $a_n$, conclude that the following series diverges: $$ \sum_{n=1}^{\infty} \frac{1}{\sqrt[n]{n+\alpha}}. $$
Explanation of Method
The key fact here is that the $n$-th root of any quantity growing at most polynomially in $n$ tends to $1$ as $n\to\infty$, because logarithmic growth (from taking the log of a polynomial) is dominated by the linear growth of $n$ in the denominator of the exponent. Applying this to $(n+\alpha)^{1/n}$ shows it tends to $1$, so its reciprocal — the general term of the series — tends to $1$ as well, not $0$, triggering the Divergence Test.
- Evaluate $\lim_{n\to\infty}(n+\alpha)^{1/n}$ via logarithms and L'Hôpital's Rule.
- Invert the result to get $\lim_{n\to\infty} a_n$.
- Apply the Divergence Test to conclude.
Step 1 · Identify the general term
$$ a_n = \frac{1}{\sqrt[n]{n+\alpha}} = \frac{1}{(n+\alpha)^{1/n}}. $$
Step 2 · Evaluate $\lim_{n\to\infty}(n+\alpha)^{1/n}$
Taking logarithms, $\ln\left[(n+\alpha)^{1/n}\right] = \dfrac{\ln(n+\alpha)}{n}$, which has the indeterminate form $\frac\infty\infty$. Applying L'Hôpital's Rule:
$$ \lim_{n\to\infty} \frac{\ln(n+\alpha)}{n} = \lim_{n\to\infty} \frac{1/(n+\alpha)}{1} = 0. $$
Since the logarithm tends to $0$ and $\exp$ is continuous:
$$ \lim_{n\to\infty} (n+\alpha)^{1/n} = e^0 = 1. $$
Step 3 · Calculate the limit of $a_n$ and apply the divergence test
$$ \lim_{n\to\infty} a_n = \frac{1}{\lim_{n\to\infty}(n+\alpha)^{1/n}} = \frac{1}{1} = 1 \neq 0. $$
By the Divergence Test, since the general term does not tend to $0$, the series $\sum_{n=1}^{\infty} \frac{1}{\sqrt[n]{n+\alpha}}$ diverges — for every real value of $\alpha$.
Summary Table
| Quantity | Value |
|---|---|
| $\lim_{n\to\infty} (n+\alpha)^{1/n}$ | $1$ |
| $\lim_{n\to\infty} a_n$ | $1$ |
| Conclusion | Diverges (Divergence Test), for any real $\alpha$ |
Additional Notes
- This result generalizes: $\lim_{n\to\infty} b_n^{1/n} = 1$ for essentially any sequence $b_n$ that grows at most exponentially (in particular for any polynomial or even sub-exponential growth in $n$), since $\ln(b_n)/n \to 0$ whenever $\ln(b_n)$ grows slower than linearly.
- The value of $\alpha$ has no effect on the conclusion: shifting $n$ by any fixed constant does not change the $n$-th root's limiting behavior, since the shift becomes negligible relative to $n$ as $n\to\infty$.
- This is a useful companion fact to Problem 16: while $(n!)^{1/n}$ grows linearly (like $n/e$) because $n!$ grows super-exponentially, $(n+\alpha)^{1/n}$ stays bounded (tending to $1$) because $n+\alpha$ only grows linearly — the contrast highlights how sensitive $n$-th-root asymptotics are to the growth rate of the base.
Result
The series diverges for every real $\alpha$ because: