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Thread: How to solve two series to figure it out the convergence / divergence

  1. #1
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    Post How to solve two series to figure it out the convergence / divergence

    Hello, guys!

    I have two series: $\displaystyle \sum_{n=1}^{\infty}(-1)^{n}\dfrac{3n-1}{n^2 + n}$

    My question is how to figure it out their convergence / divergence.
    Last edited by lebdim; Oct 21st 2018 at 05:24 AM.
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  2. #2
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    Re: How to solve two series to figure it out the convergence / divergence

    The second one is $$\sum_{n=1}^{\infty}\dfrac{\ln\left(\dfrac{n+3}{n }\right)}{3 \cdot 2^{n}}$$.
    Last edited by lebdim; Oct 21st 2018 at 05:21 AM.
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  3. #3
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    Re: How to solve two series to figure it out the convergence / divergence

    The first is an alternating series with terms that decrease to 0. Therefore it is convergent.

    For the second series, I see that "$\displaystyle \left(\frac{1}{2}\right)^n$ and immediately think "compare it to a geometric sequence". First, write it as $\displaystyle \frac{ln\left(\frac{n+ 3}{n}\right)}{3}\left(\frac{1}{2}\right)^n$

    $\displaystyle \frac{ln\left(\frac{n+ 3}{n}\right)}{3}< 1$ if and only if $\displaystyle ln\left(\frac{n+ 3}{n}\right)< 3$ which is true if and only if $\displaystyle \frac{n+ 3}{n}< e^3$. $\displaystyle n+ 3< e^3n$ if and only if $\displaystyle n(1- e^3)< 3$ or $\displaystyle (e^3- 1)n> 3$. That is, each term is less than $\displaystyle \left(\frac{1}{2}\right)^n$. The series has the convergent geometric series as upper bound so converges.
    Last edited by HallsofIvy; Oct 21st 2018 at 06:41 AM.
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    Re: How to solve two series to figure it out the convergence / divergence

    @HallsofIvy, can you show that first one is decrease to $0$.
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    Re: How to solve two series to figure it out the convergence / divergence

    Quote Originally Posted by lebdim View Post
    @HallsofIvy, can you show that first one is decrease to $0$.
    Surely you see that $\displaystyle \mathop {\lim }\limits_{n \to \infty } \dfrac{{3n - 1}}{{{n^2} + n}} = 0$ is true?

    Can you use a derivative to show that $\dfrac{3x-1}{x^2+x}$ is a decreasing function?
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