Question

show that the lim |xn+1-xn|=0

Answer 1

\[\lim_{n\to?}|x^{n+1}-x^n|=0\]

Answer 2

they're sequences

Answer 3

what is it the limit of? n approaches ?

Answer 4

it doesn't say in the question, I'll assume infinity

Answer 5

can you post a screen shot? this question is totally incomplete

Answer 6

ok give me a minute

Answer 7

oh wait I forgot it was part of a question

Answer 8

\[\lim \left|x_{n+1}-x_n\right| = \lim \left|\sum\limits_{k=1}^{n+1}\frac{1}{k} - \sum\limits_{k=1}^n\frac{1}{k}\right|\]

Answer 9

split the first sum and cancel it with the second sum

Answer 10

Okay dumb moment.. the first sum wouldn't be 1/2-1?

Answer 11

we don't need to calculate the sum
just notice that \(\large x_{n+1} - x_n\) gives you the (n+1)th term of the series

Answer 12

\[\begin{align}

\lim \left|x_{n+1}-x_n\right| &= \lim \left|\sum\limits_{k=1}^{n+1}\frac{1}{k} - \sum\limits_{k=1}^n\frac{1}{k}\right| \\~\\
&=\lim \left|\frac{1}{n+1}+\sum\limits_{k=1}^{n}\frac{1}{k} - \sum\limits_{k=1}^n\frac{1}{k}\right| \\~\\
&=\lim \left|\frac{1}{n+1}\right| \\~\\
\end{align}\]

Answer 13

take the limit

Answer 14

Ahhhhh, okay perfect! thanks!

Answer 15

lim is interpreted as limit of the sequence
it is assumed that n tends to +infinity