does this series diverge or converge?

- El_Arrow

does this series diverge or converge?

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- El_Arrow

|dw:1447129135894:dw|

- El_Arrow

and what test would you use?

- SolomonZelman

You can use the integral test, to show divergence, but don't you know that Harmonic Series diverges?

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## More answers

- El_Arrow

yeah

- El_Arrow

so you can use the comparison test and compare it to |dw:1447129370418:dw|

- El_Arrow

right?

- SolomonZelman

Yes, or you know anyway that:
\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ 1}{n+1}= \sum_{ n=2 }^{ \infty } ~ \frac{ 1}{n}}\)

- SolomonZelman

(( Harmonic series will diverge regardless of what term you start from ))

- El_Arrow

right but its not true though look |dw:1447129506901:dw|

- SolomonZelman

n+1, from n=1 and on... (2, 3, 4, 5, ....)
n, from n=2 and on... (2, 3, 4, 5, ...)
are same

- SolomonZelman

Also,
1/(n+1) < 1/n
Because you are dividng by a smaller value on the right side, and thus get a greater result on the right and smaller result on the left.

- anonymous

prove to yourself that if you have a rational expression, in order for the sum to converge the degree of the denominator must be larger than the degree of the numerator by MORE than one

- SolomonZelman

The Harmonic Series Diverges.
Integral Test:
\(\large\color{black}{\displaystyle\int\limits_{1}^{\infty}\frac{1}{x+1} ~dx=\ln(\infty+1)-\ln(1+1)=\bf Diverges}\)

- anonymous

that is called the "eyeball" test
you just look at it and see

- El_Arrow

oh okay

- El_Arrow

but if its convergent the sign would go to the left >?

- SolomonZelman

What>?

- SolomonZelman

Convergent, for a series means that the sum of the series is defined.
That it has some limit.

- SolomonZelman

Divergent, for a series means that the magnitude of the sum grows infinitely with no stop.

- El_Arrow

for example if we have |dw:1447129953807:dw|

- El_Arrow

the one on the right converges cause of the p-series

- El_Arrow

so would it be look something like this |dw:1447130054544:dw|

- SolomonZelman

A p-series where p>1 will converge

- SolomonZelman

So both of those series you have just now mentioned with n^2 on the bottom will converge

- El_Arrow

yeah i think so

- SolomonZelman

In fact:
\(\large\color{black}{\displaystyle\sum_{n=1}^{\infty }\frac{1}{n^2}=\frac{\pi^2}{6}}\)

- El_Arrow

what?

- SolomonZelman

That p series with p=2, is equal to what I said.
(by Leonard Euler)

- SolomonZelman

\(\large\color{black}{\displaystyle\sum_{i=1}^{\infty }\frac{1}{i^2}=\frac{\pi^2}{6}}\)

- El_Arrow

ok so what do i do with it?

- SolomonZelman

That is just the fact that I shared :)

- SolomonZelman

What question are you currently doing?

- El_Arrow

i just want to know if the statement is true |dw:1447130385248:dw|

- SolomonZelman

No it is not

- SolomonZelman

What is greater?
A / 100 OR A / 1000 ?

- El_Arrow

A

- SolomonZelman

A/100
OR
A/1000 ?
I don't understand you....

- El_Arrow

my bad A/100

- SolomonZelman

Yes, because you are dividng by a smaller value, right?

- El_Arrow

yes

- SolomonZelman

Now, which is greater
1/(n^2+1) OR 1/(n^2) ?

- El_Arrow

the second one

- SolomonZelman

Yes, because you are dividng by a smaller value, thus obtaining a greater result.

- El_Arrow

so 1/(n^2) converges but 1/(n^2+1) diverges

- El_Arrow

i am using the comparison test and was wondering if that could happen? @SolomonZelman

- SolomonZelman

1/n^2 is greater than 1/(n^2+1).
So if 1/(n^2) converges, then 1/(n^2+1) will all the more so converge.

- Kainui

I think becoming comfortable with the machinery of the tests is good, but I think it's also a good idea to be able to just think critically about what it is you actually have here. It is really just the harmonic series with the index shifted and the first term removed.
\[1+\sum_{n=1}^\infty \frac{1}{n+1} =\sum_{n=1}^\infty \frac{1}{n}\]

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