Question

How do you test the series $\sum {\dfrac{{\sqrt n }}{{3n + 2}}}$ from $n$ is $[0,\infty )$ for convergence?

Answer 1

Hint : In the given question, we have been asked to test the series for convergence. So as to proceed with the subsequent question we need to know which test do we have to use. Integral test states that if $f$ is a continuous, positive and decreasing function where $f\left( n \right) = {a_n}$ on the interval $\left[ {1,{\text{ }}\infty } \right)$ . then the improper integral $\int_1^\infty {f(x)dx}$ and the infinite series $\sum\limits_{n = 1}^\infty {{a_n}}$ either both of them converge or diverge.

Complete step by step solution:
We are given,
$\sum {\dfrac{{\sqrt n }}{{3n + 2}}}$
We’ll consider the analogous function f(n)=an, and test the limit:
$\Rightarrow \int_0^\infty {f(x)dx = \mathop {\lim }\limits_{t \to \infty } } \int\limits_0^t {f(x} )\;dx < \infty \\\ \;$
In this case,
$\Rightarrow \mathop {\lim }\limits_{t \to \infty } \int_0^t {\dfrac{{\sqrt x }}{{3x + 2}}} \;dx < \infty$
With sub: $z = \sqrt x ,\;{z^2} = x,\;2zdz = dx$
We have:
$\Rightarrow \mathop {\lim }\limits_{t \to \infty } \int_0^t {\dfrac{{2{z^2}}}{{3{z^2} + 2}}} dz$
$\Rightarrow \mathop {\lim }\limits_{t \to \infty } \int_0^t {\dfrac{{(\dfrac{2}{3})(3{z^2} + 2 - 2)}}{{3{z^2} + 2}}} dz$
$\Rightarrow \mathop {\lim }\limits_{t \to \infty } \int_0^t {\dfrac{2}{3} - \dfrac{4}{{9{z^2} + 6}}} dz$
Let’s focus on the indefinite integral for the second term, ie:
$\Rightarrow \int {\dfrac{4}{{9{z^2} + 6}}} dz$
Now, we’ll use the sub: $9{z^2} = 6{\tan ^2}\theta ,\;3z = \sqrt 6 \tan \theta ,\;3dz = \sqrt 6 {\sec ^2}\theta d\theta$
This will lead us to-
$\Rightarrow \int {\dfrac{4}{{6{{\tan }^2}\theta + 6}}} \; \times \;\dfrac{{\sqrt 6 }}{3}{\sec ^2}\theta d\theta$
$\Rightarrow \dfrac{4}{{3\sqrt 6 }}\int {d\theta }$
$\Rightarrow \dfrac{4}{{3\sqrt 6 }}\arctan (\dfrac{{3z}}{{\sqrt 6 }}) + C$
Now we’ll integrate the first term and then substitute back the variable.
We’ll put $\Delta$ all together
$\Rightarrow \mathop {\lim }\limits_{x \to t} \;[\dfrac{2}{3}\sqrt x - \dfrac{4}{{3\sqrt 6 }}\arctan (\sqrt {\dfrac{{3x}}{2}} ]_0^t$
Now we know that: $\mathop {\lim }\limits_{\phi \to \infty } \arctan (\phi ) = \dfrac{\pi }{2}$ so the leading $\sqrt x$ term dominates the limit, and means that this integral does not converge. This in turn means that the series does not converge.
$\Rightarrow 1$

Note : There are few important points to be taken care of –
On the improper integral the lower limit must be the similar value that starts the given series. To know when to use integral tests, simply check if the terms are positive, decreasing and can be easily integrated. If the sequence of partial sums has limit and is finite then the series is also called convergent. If the sequence of partial sums does not have its limit or is plus or minus infinity then the series is also called divergent.