- #1

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## Homework Statement

question given is 1 + 1/1! + 1/2! + 1/3! + · · ·

the question need to be compared with another equation

so how to get the second equation?

## Homework Equations

n/a

## The Attempt at a Solution

i duno hw to startT.T

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- Thread starter pusaran
- Start date

- #1

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question given is 1 + 1/1! + 1/2! + 1/3! + · · ·

the question need to be compared with another equation

so how to get the second equation?

n/a

i duno hw to startT.T

- #2

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if youre limited to using the comparison test, just try to look for a series with terms greater than 1/n! as n tends to infinity.

examples might include a p-series of some sort: ( 1/(n^p)) where p>1. although i think that factorials (n!) always surpass all exponentials (x^n) (side note: does anyone know of a proof or theorem for that?) so you might wish to try for a different convergent series.

- #3

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is tat means there are few different equation can be use to compared tis equation?

- #4

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you're trying to prove that the series (1/n!) converges, right?

- #5

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how to noe we shud use tis equation to compare with the 1st equation?

- #6

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you'll do this by making an inequality (> or <) with the two series. im assuming you know (or they told you) that 1+1/2+1/2^2+1/2^3+... converges, right?

so once you compare them to see which one is greater than the other, you can draw your conclusion as to if it converges or not.

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- #8

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- #9

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nw,question given is 1 + 1/1! + 1/2! + 1/3! + · · ·

tis is the 1st equation

if i use 1st equation to compare with 1+1/2+1/2^2+1/2^3+... (which is converges) and 1st equation is less than 2nd equation,thn the 1st equation is converge

but

if i use the 1st equation to compare with another equation which is also converges but tis time the result is 2nd equation is less than the 1st equation,thn the 1st equation is diverge?

am i right?0.0

- #10

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your first example is correct. if the series that converges is greater than the one in question, it converges.

however, your second example is incorrect. if we compare the series in question (the "1st equation") to a series that converges, but it turns out that the 1st equation is greater than the converging one, it

think about both alternatives. you could have the "1st equation" still converge AND be greater than another converging series. or on the other hand, you could have the "1st equation" be a DIVERGENT series and STILL be greater than the convergent series. The problem is that we cannot determine what its behavior is with this certain series. another one must be found that can show it converges (i.e. one that converges itself.)

so no, if a series is greater than another convergent series, it does not necessarily mean it diverges.

however, if you showed that a series (lets stick with "1st equation") is greater than a DIVERGENT series, then you know for a fact that it diverges. for example, if we say that series A is divergent, then any series BIGGER than a divergent series could not ever converge.

- #11

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k,i get wat u mean:)

so wat about the 2nd equation,we jz use any equation we lik as 2nd equation?

so wat about the 2nd equation,we jz use any equation we lik as 2nd equation?

- #12

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wat u mean by 'one that converges itself'?

- #13

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And by that I just mean you need to keep testing series that converge.

- #14

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few more thng tat bother me

i being searching for some reference on net and i found that in chapter comparison test,the example given is the series A is compared with series B while series B is change(expand/transform/intergrate/simplify/or watever)from series A.y is tis happening?

isn't tis different frm wat u teaching me?(keep testing the series)

tis is the webpage

http://tutorial.math.lamar.edu/Classes/CalcII/SeriesCompTest.aspx

i'm not saying tat u teach me wrong,jz vr blur:)

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