Dayanara Terry

2022-07-12

Proving convergence of:

$\sum _{n=0}^{\mathrm{\infty}}\frac{\mathrm{cos}\left(n\pi \right)}{3n!+1}$

$\sum _{n=0}^{\mathrm{\infty}}\frac{\mathrm{cos}\left(n\pi \right)}{3n!+1}$

Johnathan Morse

Beginner2022-07-13Added 18 answers

$|{a}_{n}|\le {b}_{n}=\frac{1}{n!}.$

Therefore since the sum $\sum _{n=0}^{\mathrm{\infty}}{b}_{n}$ converges, the given sum converges as well.

For the amount of terms, needed for accuracy 0.01, it is better to rewrite the sum as follows:

$\sum _{n=0}^{\mathrm{\infty}}{a}_{n}=\sum Even\text{}n+\sum Odd\text{}n=\sum _{n=0}^{\mathrm{\infty}}\frac{1}{3(2n)!}+\sum _{n=0}^{\mathrm{\infty}}\frac{-1}{3(2n+1)!}=\sum _{n=0}^{\mathrm{\infty}}\frac{2n+1-1}{3(2n+1)!}=\sum _{n=0}^{\mathrm{\infty}}\frac{2n}{3(2n+1)!}=\sum _{n=0}^{\mathrm{\infty}}{a}_{n}^{\ast}$

Now all summands are positive, so the sum only increases. We need to examine the rate of change of the sum and of the summands, in order to be able to make a proper guess about N - the number of summands needed to have the needed accuracy. These rates of change are:

${\mathrm{\Delta}}_{\sum}={a}_{n}^{\ast}=\frac{2n}{3(2n+1)!}$

${\mathrm{\Delta}}_{{a}_{n}^{\ast}}={a}_{n}^{\ast}-{a}_{n-1}^{\ast}=\frac{2n}{3(2n+1)!}-\frac{2n-2}{3(2n-1)!}=\frac{2}{3}[\frac{n}{(2n+1)!}-\frac{(n-1)(2n)(2n+1)}{(2n+1)!}]=\frac{2}{3}\frac{-4{n}^{3}+4{n}^{2}+3n}{(2n+1)!}$

To be sure that the sum won't increrase with more than 0.01 after $n=N$, we can look for the n, satisfying ${\mathrm{\Delta}}_{{a}_{n}^{\ast}}\le 0.001$. By guess and check one sees that the first such n is n=4. But let's not forget that ${a}_{n}^{\ast}={a}_{2n}+{a}_{2n+1}$ and:

$\sum _{n=0}^{N}{a}_{n}^{\ast}=\sum _{n=0}^{2N}{a}_{n}$

so we have to sum up at most $2\ast 4=8$ members of $\{a{\}}_{n}$ to get accuracy up to second decimal place.

However it should be noted that this is a very rough method and higher accuracies will require much less summands than what can be estimated with this method. To achieve accuracy 0.01 are in fact needed only six summands.

Therefore since the sum $\sum _{n=0}^{\mathrm{\infty}}{b}_{n}$ converges, the given sum converges as well.

For the amount of terms, needed for accuracy 0.01, it is better to rewrite the sum as follows:

$\sum _{n=0}^{\mathrm{\infty}}{a}_{n}=\sum Even\text{}n+\sum Odd\text{}n=\sum _{n=0}^{\mathrm{\infty}}\frac{1}{3(2n)!}+\sum _{n=0}^{\mathrm{\infty}}\frac{-1}{3(2n+1)!}=\sum _{n=0}^{\mathrm{\infty}}\frac{2n+1-1}{3(2n+1)!}=\sum _{n=0}^{\mathrm{\infty}}\frac{2n}{3(2n+1)!}=\sum _{n=0}^{\mathrm{\infty}}{a}_{n}^{\ast}$

Now all summands are positive, so the sum only increases. We need to examine the rate of change of the sum and of the summands, in order to be able to make a proper guess about N - the number of summands needed to have the needed accuracy. These rates of change are:

${\mathrm{\Delta}}_{\sum}={a}_{n}^{\ast}=\frac{2n}{3(2n+1)!}$

${\mathrm{\Delta}}_{{a}_{n}^{\ast}}={a}_{n}^{\ast}-{a}_{n-1}^{\ast}=\frac{2n}{3(2n+1)!}-\frac{2n-2}{3(2n-1)!}=\frac{2}{3}[\frac{n}{(2n+1)!}-\frac{(n-1)(2n)(2n+1)}{(2n+1)!}]=\frac{2}{3}\frac{-4{n}^{3}+4{n}^{2}+3n}{(2n+1)!}$

To be sure that the sum won't increrase with more than 0.01 after $n=N$, we can look for the n, satisfying ${\mathrm{\Delta}}_{{a}_{n}^{\ast}}\le 0.001$. By guess and check one sees that the first such n is n=4. But let's not forget that ${a}_{n}^{\ast}={a}_{2n}+{a}_{2n+1}$ and:

$\sum _{n=0}^{N}{a}_{n}^{\ast}=\sum _{n=0}^{2N}{a}_{n}$

so we have to sum up at most $2\ast 4=8$ members of $\{a{\}}_{n}$ to get accuracy up to second decimal place.

However it should be noted that this is a very rough method and higher accuracies will require much less summands than what can be estimated with this method. To achieve accuracy 0.01 are in fact needed only six summands.

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