Evaluate the sum S for the series, for a constant a>0: S
Evaluate the sum $S$ for the series, for a constant $a>0$: $$ S = \sum_{n=1}^{\infty}\left(\sqrt[n]{a}-\sqrt[n+1]{a}\right). $$
Explanation of Method
The general term is already presented as a difference of consecutive values of the sequence $b_n = \sqrt[n]{a} = a^{1/n}$, so no algebraic manipulation is needed to reveal the telescoping structure — only a rewrite into exponent notation. Summing collapses everything except the first and last terms, and since $a^{1/(N+1)} \to a^0 = 1$ as $N\to\infty$ (using continuity of the exponential function), the limit exists and is finite.
- Rewrite the roots as fractional exponents, confirming the telescoping form.
- Sum the telescoping terms to get a closed form for $S_N$.
- Take the limit as $N \to \infty$, using continuity of $a^x$.
Step 1 · Recognize the telescoping form
The general term $a_n = \sqrt[n]{a} - \sqrt[n+1]{a}$ is already a difference $b_n - b_{n+1}$ with $b_n = \sqrt[n]{a}$. Rewriting with fractional exponents:
$$ a_n = a^{1/n} - a^{1/(n+1)}. $$
Step 2 · Determine the partial sum $S_N$
The $N$-th partial sum is:
$$ S_N = \sum_{n=1}^{N} \left(a^{1/n} - a^{1/(n+1)}\right). $$
Writing out the terms:
$$ S_N = \left(a^{1/1}-a^{1/2}\right) + \left(a^{1/2}-a^{1/3}\right) + \cdots + \left(a^{1/(N-1)}-a^{1/N}\right) + \left(a^{1/N}-a^{1/(N+1)}\right). $$
All intermediate terms cancel, leaving:
$$ S_N = a^{1/1} - a^{1/(N+1)} = a - a^{1/(N+1)}. $$
Step 3 · Calculate the sum $S$
The sum $S$ is the limit of $S_N$ as $N \to \infty$:
$$ S = \lim_{N \to \infty} S_N = \lim_{N \to \infty} \left(a - a^{1/(N+1)}\right). $$
Since $\dfrac{1}{N+1} \to 0$ and the exponential function $a^x$ is continuous:
$$ \lim_{N \to \infty} a^{1/(N+1)} = a^0 = 1, \qquad \text{for } a>0. $$
Therefore:
$$ S = a - 1. $$
Since this limit is finite, the series converges.
Summary Table
| Quantity | Closed form |
|---|---|
| General term $a_n$ | $a^{1/n} - a^{1/(n+1)}$ |
| Partial sum $S_N$ | $a - a^{1/(N+1)}$ |
| Sum $S$ | $a - 1$ |
Additional Notes
- If $a=1$, every term $a_n = 1^{1/n} - 1^{1/(n+1)} = 1-1 = 0$, so $S=0$, consistent with the formula $S=a-1=1-1=0$.
- For $a>1$, each term $a_n>0$ since $a^{1/n}$ decreases as $n$ increases; the sum $S=a-1$ is positive, matching this. For $0
- The convergence here relies only on the continuity of $x \mapsto a^x$ at $x=0$, not on any decay-rate argument — a simpler mechanism than the comparison tests often needed for series with more complicated general terms.
Result
The partial sum is:
The sum of the series is: