Determine if the SERIES converges or DIVERGES(III)

[shamieh]

More awesome series for you to help me with..

$$\sum^{\infty}_{n = 1} \frac{tan^{-1}n}{\sqrt{1 + n^2}}$$ =

So can i use a limit comparison test and let $$b_n$$ be $$\frac{1}{n}$$

and then use the limit comparison test and obtain 1 which is $$L > 0$$ so then since 1/n or $$b_n$$ diverges then I know that $$a_n$$ diverges by the limit comparison test ?I ended up with $$\frac{n}{\sqrt{n^2 + 1}} $$ as n --> $$\infty = 1$$

 

[Prove It]

More awesome series for you to help me with..

$$\sum^{\infty}_{n = 1} \frac{tan^{-1}n}{\sqrt{1 + n^2}}$$ =

So can i use a limit comparison test and let $$b_n$$ be $$\frac{1}{n}$$

and then use the limit comparison test and obtain 1 which is $$L > 0$$ so then since 1/n or $$b_n$$ diverges then I know that $$a_n$$ diverges by the limit comparison test ?I ended up with $$\frac{n}{\sqrt{n^2 + 1}} $$ as n --> $$\infty = 1$$

I'd say that since $\displaystyle \begin{align*} \arctan{(n)} < \frac{\pi}{2} \end{align*}$ for all n, then we have

$\displaystyle \begin{align*} \sum_{n = 1}^{\infty}{ \frac{\arctan{(n)}}{\sqrt{1 + n^2 }}} < \sum_{n = 1}^{\infty}{\frac{\frac{\pi}{2}}{\sqrt{1 + n^2}}} = \frac{\pi}{2} \sum_{n = 1}^{\infty}{\frac{1}{\sqrt{1 + n^2}}} \end{align*}$

Now if we consider $\displaystyle \begin{align*} f(x) = \frac{1}{\sqrt{1 +x^2}} = \left( 1 + x^2 \right) ^{-\frac{1}{2}} \end{align*}$, we have $\displaystyle \begin{align*} f'(x) = 2x \left[ -\frac{1}{2} \left( 1 + x^2 \right) ^{-\frac{3}{2}} \right] = - \frac{x}{\left( 1 + x^2 \right) \sqrt{1 + x^2} } \end{align*}$

Since $\displaystyle \begin{align*} f'(x) < 0 \end{align*}$ for all $\displaystyle \begin{align*} x > 0 \end{align*}$, that means the function is decreasing, and that means that the integral $\displaystyle \begin{align*} \int_0^{\infty} { \frac{1}{\sqrt{1 + x^2}} \, dx} \end{align*}$ will be an overestimate for $\displaystyle \begin{align*} \sum_{n = 1}^{\infty}{ \frac{1}{\sqrt{1 + n^2}} } \end{align*}$. So if this integral is convergent, so will be the sum.

$\displaystyle \begin{align*} \int_0^{\infty}{ \frac{1}{\sqrt{1 +x^2}} \, dx } &= \lim_{b \to \infty} { \int_0^b{ \frac{1}{\sqrt{1 + x^2}} \, dx } } \\ &= \lim_{b \to \infty}{ \left[ \arctan{(x)} \right]_0^b } \\ &= \lim_{b \to \infty}{ \left[ \arctan{(b)} - \arctan{(0)} \right] } \\ &= \lim_{b \to \infty}{ \left[ \arctan{(b)} - 0 \right] } \\ &= \lim_{b \to \infty} \left[ \arctan{(b)} \right] \\ &= \frac{\pi}{2} \end{align*}$

This integral is convergent, so $\displaystyle \begin{align*} \sum_{n = 1}^{\infty}{ \frac{1}{\sqrt{1 + n^2}} } \end{align*}$ is convergent, and thus $\displaystyle \begin{align*} \sum_{n = 1}^{\infty}{ \frac{\arctan{(x)}}{\sqrt{1 + n^2}} } \end{align*}$ is also convergent by comparison.

 

[zzephod]

I'd say that since $\displaystyle \begin{align*} \arctan{(n)} < \frac{\pi}{2} \end{align*}$ for all n, then we have

...
This integral is convergent, so $\displaystyle \begin{align*} \sum_{n = 1}^{\infty}{ \frac{1}{\sqrt{1 + n^2}} } \end{align*}$ is convergent, and thus $\displaystyle \begin{align*} \sum_{n = 1}^{\infty}{ \frac{\arctan{(x)}}{\sqrt{1 + n^2}} } \end{align*}$ is also convergent by comparison.

But the series is not convergent. As $$ n \to \infty$$ we have [math]\arctan(n) \to \pi/2[/math] and $$\frac{1}{\sqrt{1+n^2}} \sim \frac{1}{n}$$.

Hence for large $$n$$ the terms behave like $$\frac{\pi}{2n}$$ and so the series diverges.The problem is with your integral:

$$\int_0^b \frac{1}{\sqrt{1+x^2}} = \mathrm{asinh}\left( b\right) \ne \arctan(b)$$

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[Prove It]

But the series is not convergent. As $$ n \to \infty$$ we have [math]\arctan(n) \to \pi/2[/math] and $$\frac{1}{\sqrt{1+n^2}} \sim \frac{1}{n}$$.

Hence for large $$n$$ the terms behave like $$1/n$$ and so the series diverges.The problem is with your integral:

$$\int_0^b \frac{1}{\sqrt{1+x^2}} = \mathrm{asinh}\left( b\right) \ne \arctan(b)$$

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Oops, that's where my mistake is, thank you :)