Question

help!!

Answer 1

show that the sequence \[X _{n}= \sum_{k=1}^{n} \frac{ 1}{ k } -Log(n+1) \]

where, n is grater and equal than 1

Answer 2

wt is ur question

Answer 3

i need to show that the sequence is convergent. Can anyone show me which way do i need to follow..

Answer 4

Is that \(-\log(k+1)\)?

Answer 5

no, log(n+1)

Answer 6

OK. Did you try to see the integral?

Answer 7

no, but i think limit would be helpful. but i am not sure.. what is your idea ?

Answer 8

Do you know the integral test for convergence?

If an integral exists with the same limits, then a sum does too.

Answer 9

Wait, is the -log(n+1) part inside or outside of the summation?

Answer 10

OP says it's not \(\log(k + 1)\), so must be outside.

Answer 11

Well, not necessarily. It would just be trickier lol.

Answer 12

Oh, yes. lol

Answer 13

what is integral boundary? 1 to n ? or 1 to infinite ?

Answer 14

If you want to show that it is convergent, 1 to infinity.

Answer 15

so what about log(n+1) part.. if n goes to infinite then, log term goes to infinite as well ..

Answer 16

That's what I was thinking.

Answer 17

also, sum part is divergent as i know, if n goes to infinite.

Answer 18

Yes, that's why I think that the \(\log(n + 1)\) is outside. It's giving me the right result.

Answer 19

so,it is weird but, that two divergent terms cancel each other,, ? :S

Answer 20

If you draw a picture it should be clear why they "cancel".

Answer 21

\[-\log{(n + 1)} + \int_{1}^{n } \dfrac{1}{n}dn\]\[-\log(n + 1) + \log(n) = \log(n/n+1)\]\[n \to \infty \implies \log(1) = 0\]

Answer 22

Whoops.

Answer 23

\[\int_{1}^{n} \dfrac{1}{k}dk\]

Answer 24

Yes, that's it. lol.

Answer 25

Here's a picture of your series. The parts that are above the curve start to decrease.\[\sum \dfrac{1}{k} = \int \dfrac{1}{k}dk + \sum \rm parts ~above~the~curve\]
https://upload.wikimedia.org/wikipedia/commons/6/67/Integral_Test.svg

Answer 26

It's clear that as you approach infinity, the area above the curve will be zero. So the sum of the "parts above the curve" converges.

Answer 27

so both two terms are convergences .. ?

Answer 28

Actually if you draw it like this: then you can definitely say \[\int\limits_1^n \frac{dk}{k} \ge\sum_1^n\frac{1}{k}\] then just subtract off ln(n+1) from both sides and take the limit like you just did. Sorry to interrupt!