# Use the comparison or limit comparison test to decide if the

Use the comparison or limit comparison test to decide if the following series converge.
$\sum _{n=1}^{\mathrm{\infty }}\frac{4-\mathrm{sin}n}{{n}^{2}+1}$
For each series which converges, give an approximation of its sum, together with an error estimate, as follows. First calculate the sum ${s}_{5}$ of the first 5 terms, then estimate the "tail" $\sum _{n=6}^{\mathrm{\infty }}{a}_{n}$ by comparing it with an appropriate improper integral or geometric series.
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Timothy Wolff
Given series is $\sum _{n=1}^{\mathrm{\infty }}\frac{4-\mathrm{sin}n}{{n}^{2}+1}$
Since $\frac{4-\mathrm{sin}n}{{n}^{2}+1}<\frac{3}{{n}^{2}}$ (since the maximum value of $\mathrm{sin}n$ is 1)
And $\sum _{n=1}^{\mathrm{\infty }}\frac{3}{{n}^{2}}$ is convergent
Since $\sum _{n=1}^{\mathrm{\infty }}\frac{4-\mathrm{sin}n}{{n}^{2}+1}<\sum _{n=1}^{\mathrm{\infty }}\frac{3}{{n}^{2}}$ and $\sum _{n=1}^{\mathrm{\infty }}\frac{3}{{n}^{2}}$ is convergent
By comparison test $\sum _{n=1}^{\mathrm{\infty }}\frac{4-\mathrm{sin}n}{{n}^{2}+1}$ is convergent
###### Not exactly what you’re looking for?
psor32
Use the limit test to determine the convergence of the series:
$\sum _{n}^{\mathrm{\infty }}\frac{4-\mathrm{sin}\left(n\right)}{{n}^{2}+1}$
Recall the statement of the limit test.
The convergence of a series $\sum _{n}^{\mathrm{\infty }}{a}_{n}$ can be determined by examining the quantity $\rho =\underset{n\to \mathrm{\infty }}{lim}{a}_{n}$ using the following comparisons:
If $|\rho |>0$ or if ρ is undefined, the series diverges.
If $\rho =0$, the limit test is inconclusive.
Check the divergence of the series by computing the limit of the summand.
Take the limit of the summand as n approaches $\mathrm{\infty }$:
$\underset{n\to \mathrm{\infty }}{lim}\frac{4-\mathrm{sin}\left(n\right)}{{n}^{2}+1}=0$
INTERMEDIATE STEPS:
Find the following limit:
$\underset{n\to \mathrm{\infty }}{lim}\frac{4-\mathrm{sin}\left(n\right)}{{n}^{2}+1}$
A bounded function times one that approaches 0 also approaches 0.
Since $3\le 4-\mathrm{sin}\left(n\right)\le 5$ for all n, and since $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{n}^{2}+1}=0$, we have that $\underset{n\to \mathrm{\infty }}{lim}\frac{4-\mathrm{sin}\left(n\right)}{{n}^{2}+1}=0:$
If the limit of the summand as the iterator n goes to $\mathrm{\infty }$ is nonzero, then the series must diverge.
Since the limit is equal to 0, the limit test is inconclusive:
Answer: The limit test is inconclusive.
user_27qwe