Investigate the convergence of the series: sum from n=1
Investigate the convergence of the series: $$ \sum_{n=1}^{\infty} \frac{(\alpha-1)^n \alpha^n}{\left(2^n+1\right) \ln (n+1)}. $$
Explanation of Method
Combining $(\alpha-1)^n\alpha^n=[\alpha(\alpha-1)]^n=\beta^n$ turns this into a ratio-test problem: since $2^n+1\sim2^n$ for large $n$, the ratio test's limit works out to $\frac{|\beta|}2$, and solving $|\beta|<2$ (i.e. $|\alpha^2-\alpha|<2$) gives a clean interval for $\alpha$. Since $\alpha^2-\alpha$ is a parabola, this quadratic inequality translates directly into a bounded range of $\alpha$.
- Simplify the term using $\beta=\alpha(\alpha-1)=\alpha^2-\alpha$.
- Apply the ratio test, obtaining limit $|\beta|/2$.
- Solve $|\alpha^2-\alpha|<2$ and check the boundary values.
Step 1 · Simplify the term
$(\alpha-1)^n\alpha^n=[\alpha(\alpha-1)]^n=\beta^n$ with $\beta=\alpha^2-\alpha$, so $a_n=\dfrac{\beta^n}{(2^n+1)\ln(n+1)}$.
Step 2 · Apply the ratio test
$$ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = |\beta|\cdot\lim_{n\to\infty}\frac{2^n+1}{2^{n+1}+1}\cdot\frac{\ln(n+1)}{\ln(n+2)} = |\beta|\cdot\frac12\cdot1 = \frac{|\beta|}2. $$
Step 3 · Solve the inequality and check boundaries
Convergence requires $\frac{|\beta|}2<1$, i.e. $|\alpha^2-\alpha|<2$, i.e. $-2<\alpha^2-\alpha<2$. The right inequality $\alpha^2-\alpha-2<0$ factors as $(\alpha-2)(\alpha+1)<0$, giving $-1<\alpha<2$; the left inequality $\alpha^2-\alpha+2>0$ holds always (discriminant negative). So convergence for $-1<\alpha<2$. At the endpoints $\alpha=-1$ or $\alpha=2$: $\beta=2$, so $a_n\approx\dfrac1{\ln(n+1)}$, and $\sum\dfrac1{\ln(n+1)}$ diverges (comparable to the divergent series in Problem 95 at the boundary). At $\alpha=0$ or $\alpha=1$: $\beta=0$, series trivially converges to $0$.
Summary Table
| Condition | Conclusion |
|---|---|
| $-1<\alpha<2$ | Converges |
| $\alpha=-1$ or $\alpha=2$ | Diverges ($\sim\sum1/\ln(n+1)$) |
| $\alpha\le-1$ or $\alpha\ge2$ | Diverges |
Additional Notes
- The parabola $\beta=\alpha^2-\alpha$ has minimum $-\frac14$ at $\alpha=\frac12$ and is zero at $\alpha=0,1$, so throughout the interior of the convergence interval $(-1,2)$, $\beta$ ranges continuously between $-\frac14$ and $2$ (exclusive) — the endpoints of the $\alpha$-interval are exactly where $|\beta|$ first reaches the critical value $2$.
- The two boundary values $\alpha=-1$ and $\alpha=2$ give the identical value $\beta=2$ (since $(-1)(-2)=2$ and $2\cdot1=2$), so both endpoints share the same divergence mechanism — a nice symmetry stemming from the parabola's shape.
- This problem is structurally similar to Problem 76's $\sum n^3/(a+1/n)^n$ in that both use the ratio test to reduce a parametrized series to a simple inequality on a combined parameter, though here the combination $\alpha(\alpha-1)$ is quadratic rather than a simple reciprocal.
Result
The series converges if and only if $-1<\alpha<2$: