Question

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

Answer 1

Answer:

-45

Step-by-step 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.

Answer 2

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

$S_N=\displaystyle\sum_{n=1}^N\bigg(\tan^{-1}\sqrt n-\tan^{-1}\sqrt{n+1}\bigg)$

$S_N=\bigg(\tan^{-1}\sqrt1-\tan^{-1}\sqrt2\bigg)+\bigg(\tan^{-1}\sqrt2-\tan^{-1}\sqrt3\bigg)+\bigg(\tan^{-1}\sqrt3-\tan^{-1}\sqrt4\bigg)+\cdots+\bigg(\tan^{-1}\sqrt N-\tan^{-1}\sqrt{N+1}\bigg)$

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

$S_N=\tan^{-1}\sqrt1-\tan^{-1}\sqrt{N+1}$

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

$\displaystyle\sum_{n=1}^\infty\bigg(\tan^{-1}\sqrt n-\tan^{-1}\sqrt{n+1}\bigg)=\lim_{N\to\infty}S_N=\tan^{-1}1-\lim_{N\to\infty}\tan^{-1}\sqrt{N+1}$

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

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