Question

determine if the series is convergent or divergent

Answer 1

\[\sum_{n=1}^{\infty}(-1)^n \frac{ \sqrt{n} }{ 3+4\sqrt{n} }\]

Answer 2

Check the limit
\[\lim_{n\to\infty}\frac{\sqrt n}{3+4\sqrt n}\]

Answer 3

is the limit 1?

Answer 4

because \[\frac{ \sqrt{n} }{ \sqrt{n} }\]

Answer 5

and how does the alternating series affect this?

Answer 6

No, in this case you have \(\dfrac{1}{4}\). You can determine this by comparing the coefficients of the \(\sqrt n\) terms in the numerator and denominator.
\[\frac{\sqrt n}{3+4\sqrt n}=\frac{1}{\frac{3}{\sqrt n}+4}\to\dfrac{1}{4}\text{ as }n\to\infty\]
The fact that this series is alternating signs does not affect this conclusion.
\[(-1)^n\frac{1}{\frac{3}{\sqrt n}+4}\to\pm\dfrac{1}{4}\text{ as }n\to\infty\]
(which means the limit doesn't exist).

Answer 7

In either case, you have that for a series \(\sum a_n\), \(a_n\not\to0\). What does this tell you about the series?

Answer 8

the series is divergent?

Answer 9

since the limit is not zero

Answer 10

thank you

Answer 11

Correct, you're welcome

Answer 12

truly appreciate your help and i hope to see you again on another problem