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Thread: How can i check if this sum is converges?

  1. #1
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    How can i check if this sum is converges?

    Hello everybody,
    How can i check if this sum is converges?
    How can i check if this sum is converges?-ques.png
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  2. #2
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    Re: How can i check if this sum is converges?

    Hey MrMath.

    Hint - Check the denominator in terms of the power and make a deduction with some kind of p-series.
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  3. #3
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    Re: How can i check if this sum is converges?

    Quote Originally Posted by chiro View Post
    Hey MrMath.

    Hint - Check the denominator in terms of the power and make a deduction with some kind of p-series.
    Can you explain more?
    I tried to do it with convergence test but i didn't success. i will be happy if you can present a full solution.
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  4. #4
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    Re: How can i check if this sum is converges?

    \sum^\infty_{n=1} \frac{40n}{(n+3)(n+4)(n+5)}=40\cdot \sum^\infty_{n=1} \frac{n}{(n+3)(n+4)(n+5)}

    Let \sum a_n, \sum b_n be two positive sequences such that a_n\leq b_n for all n.

    If \sum b_n converges, so does \sum a_n. If \sum a_n diverges, so does \sum b_n.

    \sum^\infty_{n=1}\frac{n}{(n+3)(n+4)(n+5)} \leq \sum^\infty_{n=1} \frac{n}{n^3} = \sum^\infty_{n=1} \frac{1}{n^2}

    We now have a series of the form \sum^\infty_{n=1}\frac{1}{n^p}. Can you take it from here?

    -Andy
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  5. #5
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    Re: How can i check if this sum is converges?

    Quote Originally Posted by abender View Post
    \sum^\infty_{n=1} \frac{40n}{(n+3)(n+4)(n+5)}=40\cdot \sum^\infty_{n=1} \frac{n}{(n+3)(n+4)(n+5)}

    Let \sum a_n, \sum b_n be two positive sequences such that a_n\leq b_n for all n.

    If \sum b_n converges, so does \sum a_n. If \sum a_n diverges, so does \sum b_n.

    \sum^\infty_{n=1}\frac{n}{(n+3)(n+4)(n+5)} \leq \sum^\infty_{n=1} \frac{n}{n^3} = \sum^\infty_{n=1} \frac{1}{n^2}

    We now have a series of the form \sum^\infty_{n=1}\frac{1}{n^p}. Can you take it from here?

    -Andy
    Yes. THANK YOU!!
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  6. #6
    Senior Member zzephod's Avatar
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    Re: How can i check if this sum is converges?

    Observe that for $n>0$: $$\large \frac{40n}{(n+3)(n+4)(n+5)}< \frac{40n}{n^3}=\frac{40}{n^2},$$ then:
    $$S_N=\sum_{n=1}^{N}\frac{40n}{(n+3)(n+4)(n+5)}< \sum_{n=1}^{N}\frac{40}{n^2},$$ so $S_N$ is an increasing sequence bounded above by $K=\sum_{n=1}^{\infty}\frac{40}{n^2}$ (which is convergent with sum $\pi^2/6$). An increasing sequence bounded above converges...
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