Question

Can someone please help me with this question? I REALLY need help.

What is the value of
`$ \sum_(n=1)^\infty (\tan^(-1)√(n)-\tan^(-1)√(n+1))$`? The answer should be in radians.

Answer 1

The sum telescopes. Consider the N-th partial sum of the series,

`S_N=\displaystyle\sum_(n=1)^N\bigg(\tan^(-1)\sqrt n-\tan^(-1)√(n+1)\bigg)`

`S_N=\bigg(\tan^(-1)\sqrt1-\tan^(-1)\sqrt2\bigg)+\bigg(\tan^(-1)\sqrt2-\tan^(-1)\sqrt3\bigg)+\bigg(\tan^(-1)\sqrt3-\tan^(-1)\sqrt4\bigg)+\cdots+\bigg(\tan^(-1)\sqrt N-\tan^(-1)√(N+1)\bigg)`

Notice how
`\tan^(-1)\sqrt2` is added, then immediately subtracted. The same goes for
`\tan^(-1)\sqrt3`, then
`\tan^(-1)\sqrt4`, and so on, up to
`\tan^(-1)\sqrt N`. This leaves us with

`S_N=\tan^(-1)\sqrt1-\tan^(-1)√(N+1)`

The value of the given sum is obtained as
`N\to\infty`. We get

`\displaystyle\sum_(n=1)^\infty\bigg(\tan^(-1)\sqrt n-\tan^(-1)√(n+1)\bigg)=\lim_(N\to\infty)S_N=\tan^(-1)1-\lim_(N\to\infty)\tan^(-1)√(N+1)`

To compute the limit, recall that
`\tan^(-1)x` has a range of
`\left(-\frac\pi2,\frac\pi2\right)`; in particular,
`\tan^(-1)x\to\frac\pi2` as
`x\to\infty`. So we have

`\displaystyle\sum_(n=1)^\infty\bigg(\tan^(-1)\sqrt n-\tan^(-1)√(n+1)\bigg)=\frac\pi4-\frac\pi2=\boxed{-\frac\pi4}`

Answer 2

-45

Explanation:

The question is asking for the sum of the sequence. Therefore, there is no such thing as a "radians" answer.

When you evaluate the sum, you should get -45 as your answer. To find so, you simply plug in 1 as n and other numbers then add the result.