Given a>0, find the radius of convergence, examine
Given $a>0$, find the radius of convergence, examine the endpoints, and find the sum of $\sum_{n=1}^{\infty}e^{-na}\sinh(na)\,x^n$.
Explanation of Method
Expanding $\sinh(na)=\tfrac12(e^{na}-e^{-na})$ turns the coefficient $e^{-na}\sinh(na)$ into $\tfrac12(1-e^{-2na})$, which is bounded and tends to the nonzero constant $\tfrac12$ as $n\to\infty$. A coefficient with a finite nonzero limit gives radius of convergence exactly $1$, regardless of $a$; the sum itself splits cleanly into two geometric series.
- Simplify the coefficient using the exponential definition of $\sinh$.
- Apply the root test to find $R$, using that the coefficient converges to a finite nonzero limit.
- Split the sum into two geometric series and combine into a single closed form.
Step 1 · Simplify the coefficient
$$ e^{-na}\sinh(na) = e^{-na}\cdot\frac{e^{na}-e^{-na}}2 = \frac{1-e^{-2na}}2. $$ As $n\to\infty$ (and $a>0$ fixed), this coefficient increases monotonically to $\tfrac12$. (Numerical check at $a=1$: the coefficient is $0.432$ at $n=1$, $0.500$ already by $n=5$, and indistinguishable from $\tfrac12$ for $n\ge10$.)
Step 2 · Radius of convergence and endpoints
Since $\left(\tfrac{1-e^{-2na}}2\right)^{1/n}\to1$ (any coefficient with a finite nonzero limit contributes a factor of $1$ to the root test), the radius is $$ R=1, $$ independent of $a$. At $x=1$: the general term tends to $\tfrac12\ne0$, so the series diverges. At $x=-1$: the terms alternate but their magnitude still tends to $\tfrac12\ne0$, so the necessary condition for convergence fails and the series diverges there as well. Both endpoints diverge.
Step 3 · Closed-form sum
$$ \sum_{n=1}^{\infty}\frac{1-e^{-2na}}2 x^n = \frac12\sum_{n=1}^{\infty}x^n - \frac12\sum_{n=1}^{\infty}(e^{-2a}x)^n = \frac12\left(\frac x{1-x} - \frac{e^{-2a}x}{1-e^{-2a}x}\right), \qquad |x|<1. $$ Combining over a common denominator, the numerator collapses to $\tfrac12x(1-e^{-2a}) = xe^{-a}\sinh a$: $$ \sum_{n=1}^{\infty}e^{-na}\sinh(na)\,x^n = \frac{xe^{-a}\sinh a}{(1-x)(1-e^{-2a}x)}. $$ (Verified numerically at $a=1,x=0.5$: closed form gives $0.463711$, matching a 300-term partial sum to 6 decimal places.)
Summary Table
| Property | Value |
|---|---|
| Radius | $R=1$ (independent of $a$) |
| At $x=\pm1$ | diverges (both endpoints) |
| Sum | $\dfrac{xe^{-a}\sinh a}{(1-x)(1-e^{-2a}x)}$ |
Additional Notes
- Because the coefficient $\tfrac12(1-e^{-2na})$ tends to a nonzero constant rather than decaying, the radius of convergence is fixed at $1$ for every $a>0$ — a useful contrast with parametric series like Problems 311–312, where the parameter directly reshapes the radius itself.
- Divergence at both endpoints here mirrors Problem 311's $\alpha=1,x=+1/(ae)$ case, but for a stronger reason: there the term merely decayed too slowly ($\sim1/\sqrt n$), while here the term does not decay to $0$ at all.
- The two-geometric-series decomposition technique — splitting a bounded, exponentially-approaching-a-constant coefficient into a sum and difference of pure geometric series — is the same device used for the generating functions in Problems 262–263 and 315.