Evaluate the sum S for the series: S = sum from n=1
Evaluate the sum $S$ for the series: $$ S = \sum_{n=1}^{\infty} \arctan \frac{1}{2n^2}. $$
Explanation of Method
The arctangent subtraction formula, $\arctan x - \arctan y = \arctan\left(\frac{x-y}{1+xy}\right)$, suggests looking for $x$ and $y$ whose combination reproduces $\frac{1}{2n^2}$. Trying $x=\frac1{2n-1}$ and $y=\frac1{2n+1}$ works out exactly, which converts each term of the series into a difference of consecutive arctangent values — a telescoping structure — so the partial sums collapse and the limit can be evaluated using continuity of $\arctan$.
- Find $x,y$ such that $\arctan x - \arctan y = \arctan\frac{1}{2n^2}$, using the subtraction formula.
- Sum the resulting telescoping terms to get $S_N$.
- Take the limit as $N \to \infty$, using continuity of $\arctan$.
Step 1 · Decompose the general term
The general term is $a_n = \arctan \dfrac{1}{2n^2}$. Using the subtraction formula $\arctan x-\arctan y = \arctan\left(\frac{x-y}{1+xy}\right)$, try $x=\frac1{2n-1}$, $y=\frac1{2n+1}$:
$$ x-y = \frac{1}{2n-1}-\frac1{2n+1} = \frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)} = \frac{2}{4n^2-1}, $$
$$ 1+xy = 1+\frac{1}{(2n-1)(2n+1)} = \frac{(4n^2-1)+1}{4n^2-1} = \frac{4n^2}{4n^2-1}. $$
So the argument of the resulting arctangent is:
$$ \frac{x-y}{1+xy} = \frac{2/(4n^2-1)}{4n^2/(4n^2-1)} = \frac{2}{4n^2} = \frac{1}{2n^2}. $$
Thus:
$$ a_n = \arctan\left(\frac{1}{2n-1}\right) - \arctan\left(\frac{1}{2n+1}\right), $$
a difference $b_n - b_{n+1}$ with $b_n = \arctan\left(\frac{1}{2n-1}\right)$.
Step 2 · Determine the partial sum $S_N$
This is a telescoping series. The $N$-th partial sum is:
$$ S_N = \sum_{n=1}^{N}\left[\arctan\left(\frac1{2n-1}\right)-\arctan\left(\frac1{2n+1}\right)\right]. $$
Writing out the terms:
$$ S_N = \left[\arctan(1)-\arctan\tfrac13\right] + \left[\arctan\tfrac13-\arctan\tfrac15\right] + \cdots + \left[\arctan\tfrac1{2N-1}-\arctan\tfrac1{2N+1}\right]. $$
After cancellation:
$$ S_N = \arctan(1) - \arctan\left(\frac{1}{2N+1}\right). $$
Step 3 · Calculate the sum $S$
The sum $S$ is the limit of $S_N$ as $N \to \infty$:
$$ S = \lim_{N\to\infty} S_N = \lim_{N\to\infty}\left[\arctan(1)-\arctan\left(\frac1{2N+1}\right)\right]. $$
Since $\dfrac{1}{2N+1}\to0$ and $\arctan$ is continuous with $\arctan(0)=0$:
$$ \lim_{N\to\infty}\arctan\left(\frac1{2N+1}\right) = 0. $$
Therefore:
$$ S = \arctan(1) - 0 = \frac{\pi}{4}. $$
Summary Table
| Quantity | Closed form |
|---|---|
| General term $a_n$ | $\arctan\frac1{2n-1}-\arctan\frac1{2n+1}$ |
| Partial sum $S_N$ | $\dfrac{\pi}{4}-\arctan\dfrac1{2N+1}$ |
| Sum $S$ | $\dfrac{\pi}{4}$ |
Additional Notes
- The key algebraic step is recognizing $\frac{1}{2n^2}$ as the output of the arctangent subtraction formula for the specific pair $\frac1{2n-1}, \frac1{2n+1}$ — an identity that must be verified (as done in Step 1), not guessed and left unchecked.
- Since $\arctan(1)=\frac\pi4$ is the first term of the telescoping chain, the final sum inherits this value exactly once the vanishing tail $\arctan\frac1{2N+1}\to0$ is subtracted off.
- All terms $a_n>0$ (since $\arctan$ is increasing and $\frac1{2n-1}>\frac1{2n+1}$), so the partial sums increase monotonically toward $\frac\pi4$ from below, mirroring the pattern seen in Problems 3 and 5.
Result
The partial sum is:
The sum of the series is: