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: $$ \sum_{n=1}^{\infty} \frac{1}{1+a^n} \qquad (a>0). $$
Explanation of Method
The behavior of $a^n$ splits into three qualitatively different regimes depending on $a$: shrinking to $0$ ($0<a<1$), staying constant ($a=1$), or blowing up ($a>1$). In the first two cases the general term tends to a nonzero constant, so the Divergence Test settles things immediately. In the third case, $a^n$ dominates the denominator for large $n$, making the general term comparable to the convergent geometric series $\sum 1/a^n$, which the limit comparison test confirms.
- Analyze the case $0<a\le1$ using the Divergence Test.
- Analyze the case $a>1$ using comparison with the geometric series $\sum 1/a^n$.
- Combine the cases into a single convergence criterion.
Step 1 · Case $0<a<1$ and $a=1$
If $0<a<1$: $a^n\to0$, so $a_n=\frac1{1+a^n}\to\frac1{1+0}=1\ne0$; by the Divergence Test the series diverges.
If $a=1$: $a_n=\frac1{1+1}=\frac12$ for every $n$, so $\sum a_n = \sum\frac12=\infty$; diverges.
Step 2 · Case $a>1$
Here $a^n\to\infty$, so for large $n$, $a_n\approx\frac1{a^n}$. Using the limit comparison test with $b_n=\frac1{a^n}$:
$$ \lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\frac{a^n}{1+a^n} = \lim_{n\to\infty}\frac{1}{a^{-n}+1} = 1. $$
Since this limit is positive and finite, and $\sum\frac1{a^n}$ is a convergent geometric series (ratio $\frac1a<1$, summing to $\frac1{a-1}$), the series $\sum\frac1{1+a^n}$ converges as well.
Step 3 · Direct comparison confirmation for $a>1$
Since $1+a^n\ge a^n$, we get $\frac1{1+a^n}\le\frac1{a^n}$, and comparison with the convergent geometric series confirms convergence directly, without needing the limit form.
Summary Table
| Value of $a$ | Convergence |
|---|---|
| $0<a<1$ | Diverges (general term $\to1\ne0$) |
| $a=1$ | Diverges (constant terms $\tfrac12$) |
| $a>1$ | Converges (comparable to $\sum 1/a^n$) |
Additional Notes
- This is a nice example combining two different tests within a single problem: the Divergence Test handles the "easy" cases where the terms don't vanish, while the limit comparison test is reserved for the case where they do — a common pattern when the series depends on a parameter.
- The threshold at $a=1$ is sharp and includes the boundary case in the divergent regime: even though $a^n$ stays bounded (equal to $1$) rather than shrinking, the terms still fail to vanish, so divergence follows the same way as for $0<a<1$.
- For $a>1$, the exact sum is not requested here, but a closed form does exist in special cases (e.g. via the Lambert series or digamma-function identities); the comparison approach used here only certifies convergence, not the value.
Result
The convergence of the series depends on $a$: