More Examples on Series

Problem 1: Test for convergence

displaymath194.

Answer: Since we have a power n in the series, we will use the Root-Test. Set

displaymath196.

We have

displaymath198.

Since

displaymath200,

and

displaymath202,

we get

displaymath204.

But,

displaymath206.

Hence,

displaymath208.

Therefore, by the Root-Test, the series

displaymath194

is divergent.

Problem 2: Test for convergence

displaymath212.

Answer: The sum of two series converges, if both of the sums converge. Hence the series

$\displaystyle \sum$$\displaystyle {\frac{(3n)!+4^{n+1}}{(3n+1)!}}$

will diverge, if we can show that

$\displaystyle \sum$$\displaystyle {\frac{(3n)!}{(3n+1)!}}$

diverges, while the series

$\displaystyle \sum$$\displaystyle {\frac{4^{n+1}}{(3n+1)!}}$

converges. Since

$\displaystyle {\frac{(3n)!}{(3n+1)!}}$ = $\displaystyle {\frac{1}{3n+1}}$,

and

$\displaystyle {\frac{1}{3n+1}}$ $\displaystyle \approx$ $\displaystyle {\frac{1}{3n}}$

and the series

$\displaystyle \sum$$\displaystyle {\frac{1}{3n}}$

diverges by the p-test, we conclude that

$\displaystyle \sum$$\displaystyle {\frac{(3n)!}{(3n+1)!}}$

diverges.

On the other hand,

$\displaystyle \sum$$\displaystyle {\frac{4^{n+1}}{(3n+1)!}}$

converges by the ratio test:

$\displaystyle \lim_{n\to\infty}^{}$$\displaystyle {\frac{\displaystyle\frac{4^{n+2}}{(3n+2)!}}{\displaystyle\frac{4^{n+1}}{(3n+1)!}
}}$ = $\displaystyle \lim_{n\to\infty}^{}$$\displaystyle {\frac{4}{3n+2}}$ = 0 < 1.

This establishes that

$\displaystyle \sum$$\displaystyle {\frac{(3n)!+4^{n+1}}{(3n+1)!}}$

diverges.

Problem 3: Test for convergence

displaymath232.

Answer: We will use the Ratio-Test (try to use the Root-Test to see how difficult it is). Set

displaymath234.

We have

displaymath236.

Algebraic manipulations give

displaymath238,

since

displaymath240.

Hence, we have

displaymath242,

which implies

displaymath244.

Since tex2html_wrap_inline246 , we conclude, from the Ratio-Test, that the series

displaymath232

is convergent.

Problem 4: Determine whether the series

displaymath250

is convergent or divergent.

Answer: Consider the function

displaymath252.

It is easy to check that f(x) is decreasing on tex2html_wrap_inline256 . Hence, for any tex2html_wrap_inline258 , we have for any tex2html_wrap_inline260,

displaymath262,

which implies

displaymath264,

that is,

displaymath266.

Using this inequality, we get

displaymath268,

since

displaymath270.

Since

displaymath272,

we deduce that the partial sums associated to the series

displaymath250

are not bounded. Therefore, the series

displaymath250

is divergent.

Remark: Note that the proof given above is the proof of the Integral-Test. In other words, we may have just used to Integral-Test to get the conclusion. Also, the series given here is part of a type of series called Bertrand series defined as

displaymath278.

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