How many terms of the series must be summed to obtain
How many terms of the series must be summed to obtain the sum with accuracy $10^{-6}$: $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n^2+1}}. $$
Explanation of Method
Since $a_n=\frac1{\sqrt{n^2+1}}$ behaves like $\frac1n$ for large $n$, this series converges only about as fast as the alternating harmonic series — meaning an extremely tight tolerance like $10^{-6}$ demands roughly a million terms, in stark contrast to the handful needed in Problems 121–122 where the terms decayed polynomially much faster.
- Verify convergence via Leibniz's test.
- Solve $\frac1{\sqrt{(N+1)^2+1}}<10^{-6}$ for $N$.
- Report the required number of terms.
Step 1 · Verify convergence
$a_n=\frac1{\sqrt{n^2+1}}>0$, decreasing (since $\sqrt{n^2+1}$ increases), and $a_n\to0$. By Leibniz's test the series converges.
Step 2 · Solve for the required $N$
$$ \frac1{\sqrt{(N+1)^2+1}} < 10^{-6} \iff \sqrt{(N+1)^2+1}>10^6 \iff (N+1)^2+1>10^{12}. $$
Since the $+1$ is negligible against $10^{12}$: $(N+1)^2>10^{12}-1\approx10^{12}$, so $N+1>10^6$, giving $N>999999$.
Step 3 · Report the required number of terms
At $N=1{,}000{,}000$: $a_{1000001}=\dfrac1{\sqrt{10^{12}+2\cdot10^6+2}}<\dfrac1{10^6}=10^{-6}$, confirming this is sufficient. So $N=1{,}000{,}000$ terms are needed.
Summary Table
| Quantity | Value |
|---|---|
| Required accuracy | $10^{-6}$ |
| Terms needed $N$ | $1{,}000{,}000$ |
| Asymptotic behavior | $a_n\sim1/n$ |
Additional Notes
- This is a dramatic illustration of the practical cost of slow convergence: since $a_n\sim1/n$ (the same rate as the alternating harmonic series), roughly $10^6$ terms are needed for six-digit accuracy — a general rule of thumb for "$1/n$-type" alternating series is that the number of terms needed scales like $10^{k}$ for $k$-digit accuracy.
- Compare directly with Problem 121's $1/n^4$-type series, where only 3 terms sufficed for two-digit accuracy — the difference in required terms (3 versus a million) for a comparable relative accuracy demonstrates just how sensitive alternating-series convergence speed is to the polynomial decay rate of $a_n$.
- In practice, for series this slow, one would use acceleration techniques (such as Euler transformation or the Euler–Maclaurin-style corrections from Problem 97) rather than brute-force summation of a million terms.
Result
One million terms are needed for accuracy $10^{-6}$: