Question

Converging or diverging sequence? How do you tell which is which?

Answer 1

Try checking the ratio between successive terms.

Answer 2

Substitute values

Answer 3

Another thing you might find useful: check the limit as \(n\to\infty\). If the limit is non-zero, the sequence diverges. (The converse is not true)

Answer 4

@SithsAndGiggles @cambrige I tried to substitute the values and got a = 33.33 repeating

Answer 5

Substitute which values?

Answer 6

I put 50 for n

Answer 7

start with 1

Answer 8

then 2

Answer 9

then find the ratio

Answer 10

Why stop there? Compare the \(n\)th and \(n+1\)-th terms:
\[\frac{a_{n+1}}{a_n}\to\text{what value?}\]

Answer 11

follow Siths

Answer 12

then check if ration is >1 or< 1

Answer 13

the ration seems to be greater than one @cambrige @SithsAndGiggles

Answer 14

what do I do next

Answer 15

A ratio greater than one indicates divergence. Think of the geometric sequence \(\{1,2,4,8,\ldots\}\) which has common ratio 2 (clearly greater than 1).

Answer 16

@sithsandgiggles so is my answer B? @cambrige

Answer 17

Yea, it's D

Answer 18

@Brostep0s @sithsandgiggles B or D...?

Answer 19

DDDD

Answer 20

@Brostep0s thanks

Answer 21

yeppers cx

Answer 22

\[\lim_{n\to\infty}\frac{(-1)^n(100n)}{100+n}=\pm100\lim_{n\to\infty}\frac{1}{\dfrac{100}{n}+1}=\pm100\]
The limit is non-zero, so the sequence diverges.

You arrive at the same conclusion with the ratio test:
\[\begin{align*}
\left|\frac{a_{n+1}}{a_n}\right|&=\left|\frac{\dfrac{(-1)^{n+1}(100(n+1))}{100+n+1}}{\dfrac{(-1)^n(100n)}{100+n}}\right|\\\\
&=\frac{\dfrac{100n+100}{101+n}}{\dfrac{100n}{100+n}}\\\\
&=\frac{(100n+100)(100+n)}{100n(101+n)}\\\\
&=\frac{100n^2+10100n+10000}{100n^2+10100n}\\\\
&=\frac{100+\dfrac{10100}{n}+\dfrac{10000}{n^2}}{100+\dfrac{10100}{n}}\\\\
&\to\frac{100}{100}=1\text{ as }n\to\infty
\end{align*}\]
Since the ratio is not less than 1, the sequence diverges.