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Determine whether the series

$\sum _{n=1}^{\infty \:\:}\frac{\left(-1\right)^n}{(3n)!}$

is convergent or divergent. If it is convergent, then how many terms of the series do we need in order to find the sum to within $10^{-5}$?

I'm trying to use the Alternating Series Test

$a_n=\left(-1\right)^n\:$

$b_n=\frac{1}{\left(3n\right)!}$

need to check if

$\lim _{n\to \infty }\left(\frac{1}{\left(3n\right)!}\right)$ $=\:0$

which is does because

$\frac{1->\:1}{\left(3n\right)!\:->\infty \:}\:=0$

then, need to check if $\left\{\frac{1}{\left(3n\right)!}\right\}$ is decreasing

$b_{n+1}\frac{1}{\left(3\left(n+1\right)\right)!}$ $=\:\frac{1}{(3n+3)!}$

$b_{n+1}<\:b_n$ and therefore decreasing

so the series is convergent by Alternating Series Test

Now for the other part of the question start writing out the series

$\sum _{n=1}^{\infty }\:\frac{\left(-1\right)^n}{\left(3n\right)!}\:=\:\frac{-1}{3!}+\frac{1}{6!}-\frac{1}{9!}\:$

$\frac{1}{6!}=1.388\cdot 10^{-3}$

$-\frac{1}{9!}\:=-2.755\:\cdot 10^{-6}$

since the question is within $10^{-5}$ i would think not to include $-\frac{1}{9!}$

and say i only include 2 terms. Does my work and answers look correct?

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  • $\begingroup$ Why not use the ratio test? If the lim as n goes to infinity is less than 1, then the series converges. $\endgroup$
    – dustin
    Nov 14, 2014 at 21:22
  • $\begingroup$ Is it $\dfrac{(-1)^n}{3n!}$ or $\dfrac{(-1)^n}{(3n)!}$? $\endgroup$
    – egreg
    Nov 14, 2014 at 21:25
  • $\begingroup$ does the ratio test have a remainder thm also? @dustin $\endgroup$
    – Charlene
    Nov 14, 2014 at 21:25
  • $\begingroup$ @egreg the (3n)! $\endgroup$
    – Charlene
    Nov 14, 2014 at 21:30
  • $\begingroup$ A related question. $\endgroup$
    – Lucian
    Nov 15, 2014 at 5:44

1 Answer 1

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Edit: This is an answer for the previous version, where an error was $10^{-15}$. According to a suggestion I am leaving it, as a hint, what one should do if the error is smaller than $10^{-5}$.

Near good, but in your question is $10^{-15}$. It is known, that the error is less than the firts of remaining terms, hence $(3n)!>10^{15}$, which is true from $n=6$, hence you need two additional terms.

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  • $\begingroup$ oh i'm sorry its 10^-5 $\endgroup$
    – Charlene
    Nov 14, 2014 at 21:24
  • $\begingroup$ @Charlene So it is OK, if it is (3n)!. Should I then remove my answer? $\endgroup$ Nov 14, 2014 at 21:26
  • $\begingroup$ I would maybe remove the part about $10^{-15}$, but otherwise the explanation of why this is the correct place to truncate the series is worth keeping I think. $\endgroup$
    – BaronVT
    Nov 14, 2014 at 21:29
  • $\begingroup$ Either way doesn't matter to me, thank you for catching my mistake $\endgroup$
    – Charlene
    Nov 14, 2014 at 21:31

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