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 $10^{-6}$: $$ \sum_{n=1}^{\infty} \frac{\sin n^{\circ}}{\sqrt{n}}. $$
Explanation of Method
Converting degrees to radians, $\sin n^\circ=\sin\left(\frac{n\pi}{180}\right)$, and this series is not simply alternating — it requires Dirichlet's test, with the classical closed-form sine-sum bounding the partial sums of $\sin(n\pi/180)$ by a fixed constant $M\approx115$. The Dirichlet-test tail estimate $|R_N|\lesssim M/\sqrt N$ then converts the accuracy requirement into a (very large) bound on $N$.
- Convert to radians and bound the partial sums of $\sin(n\pi/180)$.
- Verify convergence via Dirichlet's test.
- Estimate the tail and solve for the required $N$.
Step 1 · Convert and bound partial sums
With $\theta=\frac\pi{180}$, using $\sum_{n=1}^N\sin n\theta=\dfrac{\sin\left(\frac{(N+1)\theta}2\right)\sin\left(\frac{N\theta}2\right)}{\sin(\theta/2)}$:
$$ |A_N| \le \frac1{\sin(\pi/360)} = \frac1{\sin(0.5^\circ)} \approx 114.6. $$
Step 2 · Verify convergence via Dirichlet's test
With $b_n=\frac1{\sqrt n}$ decreasing to $0$, and bounded partial sums $A_N$ from Step 1, all conditions of Dirichlet's test hold, so the series converges.
Step 3 · Estimate the required $N$
Using the Dirichlet-test tail bound $|R_N|\lesssim M\cdot b_{N+1}=\dfrac{M}{\sqrt{N+1}}$ with $M\approx115$:
$$ \frac{115}{\sqrt{N+1}} < 10^{-6} \implies \sqrt{N+1} > 1.15\times10^8 \implies N+1 > 1.32\times10^{16}. $$
So approximately $N\approx1.3\times10^{16}$ terms are required — an impractically large number for direct summation.
Summary Table
| Quantity | Value |
|---|---|
| Bound on partial sums $A_N$ | $\approx115$ |
| Tail estimate | $|R_N|\lesssim115/\sqrt N$ |
| Required $N$ | $\approx1.3\times10^{16}$ |
Additional Notes
- This series converges by Dirichlet's test rather than Leibniz's test, since $\sin(n\pi/180)$ does not alternate sign every single term but instead over blocks of roughly $180$ consecutive terms (positive for $0<n<180$, negative for $180<n<360$, with period $360$) — a fundamentally different oscillation pattern from a term-by-term alternating series.
- The astronomically large term count (over $10^{16}$) reflects both the slow decay $b_n=1/\sqrt n$ (already slower than the $1/n$ case of Problem 123) and the conservative nature of the Dirichlet bound — in practice, actual cancellation is likely better than the worst-case estimate suggests, but a precise numeric answer would require computational analysis beyond hand calculation.
- This problem, together with Problem 123, illustrates a spectrum of convergence speeds for series with polynomially-decaying terms: $1/n^4$ (fast, Problem 121), $1/n$ (slow, Problem 123), and $1/\sqrt n$ (even slower, this problem) — each requiring dramatically more terms than the last for the same target accuracy.
Result
Approximately $10^{16}$ terms are required — an extremely slowly converging series: