What is the error if the sum of sum from n=1 to infinity
What is the error if the sum of $\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}n$ is replaced by the sum of the first $100$ terms?
Explanation of Method
The series is alternating with terms decreasing to zero, so the Leibniz alternating-series bound applies directly: the error is no larger than the magnitude of the first omitted term, and carries the same sign.
- Confirm the Leibniz conditions hold ($a_n=1/n$ positive, decreasing, $\to0$).
- Apply the bound $|R_N|\le a_{N+1}$ with $N=100$.
- Determine the sign and check numerically.
Step 1 · Leibniz's theorem
For an alternating series with $a_n>0$, decreasing, $a_n\to0$, the remainder after $N$ terms satisfies $|R_N|\le a_{N+1}$, with the same sign as the first omitted term. Here $a_n=1/n$ satisfies all three conditions.
Step 2 · Apply the bound
The first omitted term (after $100$ terms) is $a_{101}=\dfrac1{101}$: $$ |R_{100}| \le \frac1{101} \approx 0.0099. $$
Step 3 · Sign and numerical check
The first omitted term $a_{101}=(-1)^{100}/101=1/101>0$, so $R_{100}=S-S_{100}>0$, i.e. $S_{100}$ underestimates $S=\ln2\approx0.693147$. Numerically, $S_{100}\approx0.688172$, giving actual error $\approx0.004975<\frac1{101}$, consistent with the bound.
Summary Table
| Property | Value |
|---|---|
| Error bound | $|R_{100}|\le1/101\approx0.0099$ |
| Sign | $S_{100}<\ln2$ (underestimate) |
| Actual error | $\approx0.004975$ |
Additional Notes
- The Leibniz bound is often loose by roughly a factor of $2$, exactly as seen here: the actual error ($\approx0.00498$) is about half the bound ($\approx0.0099$).
- The alternating harmonic series converges extremely slowly — needing on the order of $1/\varepsilon$ terms for an error of $\varepsilon$ — in sharp contrast to the factorially fast convergence of the exponential series analyzed in Problem 280.
- This is the same series encountered as $\ln(1+x)$ at $x=1$ throughout Problems 261 and 272, so this error estimate directly quantifies how many terms are needed to compute $\ln 2$ to a given precision using that series.