What is the sum of the infinite telescopic series $\sum_{n=1}^\infty \arctan (n+4)-\arctan(n+2))$? Trying to find the sum of the telescopic series. Can't seem to figure it out! Please Help!
$$\sum_{1}^{\infty}(\tan^{-1}(n+4)-\tan^{-1}(n+2))$$
I've tried looking at other examples, but they're not helping.
 A: Hint.  It doesn't matter much what the function is, telescopic series always work much the same way.  For example if we have a finite sum,
$$\eqalign{\smash{\sum_{n=1}^N \bigl(f(n+4)-f(n+2)\bigr)}
  &=f(5)-f(3)+f(6)-f(4)\cr
  &\quad{}+f(7)-f(5)+f(8)-f(6)\cr
  &\quad{}+\cdots\cr
  &\quad{}+f(N+3)-f(N+1)+f(N+4)-f(N+2)\ .\cr}$$
Now if you look carefully you will see that many terms cancel and we get
$$\sum_{n=1}^N \bigl(f(n+4)-f(n+2)\bigr)
  =-f(3)-f(4)+f(N+3)+f(N+4)\ .$$
For your problem you need to replace $f$ by $\arctan$, then see what happens as $N\to\infty$.
Good luck!
A: Compute the first few terms of the sum and try if you can spot something obvious. So the sum is:
$$(\arctan(1+4) - \arctan(1+2)) + (\arctan(2+4) - \arctan(2+2)) + (\arctan(3+4) - \arctan(3+2)) + (\arctan(4+4) - \arctan(4+2)) + ...$$
You can spot that after the $5$th term cancelling begins. Note that there isn't "partner" for $\arctan(1+2)$, $\arctan(2+2)$, $\arctan(\infty +3)$ and $\arctan(\infty +4)$
So after telescoping the sum is:
$$\sum_{n=1}^{\infty} \arctan(n+4) - \arctan(n+2) = \arctan(\infty +3) + \arctan(\infty +4) - \arctan(3) - \arctan(4) = 2\arctan{\infty} - \arctan(3) - \arctan(4)$$
