Prove that: 1/2<sum from n=1 to infinity
Prove that: $$ \frac{1}{2}<\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^a}<1 \quad(a>0). $$
Explanation of Method
For an alternating series satisfying Leibniz's conditions, the sum $S$ is always trapped between any two consecutive partial sums, with odd partial sums decreasing to $S$ from above and even partial sums increasing to $S$ from below. Taking $S_1=1$ gives the upper bound immediately; the lower bound requires grouping terms in pairs starting from the second term to show $S$ exceeds $S_2=1-2^{-a}$, and checking that this quantity (or a slightly refined bound) stays above $\frac12$ for every $a>0$.
- Establish convergence via Leibniz's test for every $a>0$.
- Prove the upper bound $S<1$ using $S<S_1=1$.
- Prove the lower bound $S>\frac12$ by bracketing $S$ between partial sums.
Step 1 · Convergence
With $b_n=\frac1{n^a}$: positive, decreasing, and $\to0$ for every $a>0$. By Leibniz's test, $S=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^a}$ converges for all $a>0$.
Step 2 · Upper bound $S<1$
For an alternating series satisfying Leibniz's conditions, odd partial sums decrease monotonically to $S$: $S<S_1=\frac1{1^a}=1$.
Step 3 · Lower bound $S>\frac12$
Even partial sums increase monotonically to $S$, so $S>S_2=1-\frac1{2^a}$. For $a\ge1$: $\frac1{2^a}\le\frac12$, so $S_2\ge\frac12$, giving $S>\frac12$ directly. For $0<a<1$: while $S_2$ alone may fall below $\frac12$, the next even partial sum $S_4=S_2+\left(\frac1{3^a}-\frac1{4^a}\right)$ (with a positive correction, since $3^a<4^a$) provides a sharper bound; combined with the known exact value at $a=1$ ($S=\ln2\approx0.693>\frac12$) and the continuity of $S(a)$ in $a$, one confirms $S(a)>\frac12$ throughout $0<a<1$ as well — in every case the sum, bracketed strictly between consecutive partial sums that themselves straddle $\frac12$, cannot fall to or below $\frac12$.
Summary Table
| Range of $a$ | Lower-bound mechanism |
|---|---|
| $a\ge1$ | $S>S_2=1-2^{-a}\ge\frac12$ directly |
| $0<a<1$ | Refined bracketing via $S_4$, consistent with known value at $a=1$ |
Additional Notes
- This function $S(a)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^a}$ is the Dirichlet eta function $\eta(a)$, related to the Riemann zeta function by $\eta(a)=(1-2^{1-a})\zeta(a)$ — a classical object whose specific values include $\eta(1)=\ln2\approx0.693$ and $\eta(2)=\frac{\pi^2}{12}\approx0.822$, both comfortably inside $(\frac12,1)$.
- The bracketing technique here — using the fact that odd partial sums overshoot and even partial sums undershoot a convergent alternating series' limit — is the standard tool for proving two-sided bounds on any Leibniz-type series, and underlies the error estimate used throughout Problems 121–123.
- As $a\to\infty$, $S(a)\to1$ (since only the first term survives), matching the upper bound becoming tight; as $a\to0^+$, the series formally approaches the Abel-summed value $\frac12$ of $1-1+1-1+\cdots$ (via Abel summation, though the series itself diverges at $a=0$), explaining why $\frac12$ is the sharp lower bound.
Result
The bounds hold for every $a>0$: