El_Arrow
  • El_Arrow
does this series diverge or converge?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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El_Arrow
  • El_Arrow
|dw:1447129135894:dw|
El_Arrow
  • El_Arrow
and what test would you use?
SolomonZelman
  • SolomonZelman
You can use the integral test, to show divergence, but don't you know that Harmonic Series diverges?

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El_Arrow
  • El_Arrow
yeah
El_Arrow
  • El_Arrow
so you can use the comparison test and compare it to |dw:1447129370418:dw|
El_Arrow
  • El_Arrow
right?
SolomonZelman
  • 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
  • SolomonZelman
(( Harmonic series will diverge regardless of what term you start from ))
El_Arrow
  • El_Arrow
right but its not true though look |dw:1447129506901:dw|
SolomonZelman
  • 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
  • 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
  • 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
  • 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
  • anonymous
that is called the "eyeball" test you just look at it and see
El_Arrow
  • El_Arrow
oh okay
El_Arrow
  • El_Arrow
but if its convergent the sign would go to the left >?
SolomonZelman
  • SolomonZelman
What>?
SolomonZelman
  • SolomonZelman
Convergent, for a series means that the sum of the series is defined. That it has some limit.
SolomonZelman
  • SolomonZelman
Divergent, for a series means that the magnitude of the sum grows infinitely with no stop.
El_Arrow
  • El_Arrow
for example if we have |dw:1447129953807:dw|
El_Arrow
  • El_Arrow
the one on the right converges cause of the p-series
El_Arrow
  • El_Arrow
so would it be look something like this |dw:1447130054544:dw|
SolomonZelman
  • SolomonZelman
A p-series where p>1 will converge
SolomonZelman
  • SolomonZelman
So both of those series you have just now mentioned with n^2 on the bottom will converge
El_Arrow
  • El_Arrow
yeah i think so
SolomonZelman
  • SolomonZelman
In fact: \(\large\color{black}{\displaystyle\sum_{n=1}^{\infty }\frac{1}{n^2}=\frac{\pi^2}{6}}\)
El_Arrow
  • El_Arrow
what?
SolomonZelman
  • SolomonZelman
That p series with p=2, is equal to what I said. (by Leonard Euler)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{\displaystyle\sum_{i=1}^{\infty }\frac{1}{i^2}=\frac{\pi^2}{6}}\)
El_Arrow
  • El_Arrow
ok so what do i do with it?
SolomonZelman
  • SolomonZelman
That is just the fact that I shared :)
SolomonZelman
  • SolomonZelman
What question are you currently doing?
El_Arrow
  • El_Arrow
i just want to know if the statement is true |dw:1447130385248:dw|
SolomonZelman
  • SolomonZelman
No it is not
SolomonZelman
  • SolomonZelman
What is greater? A / 100 OR A / 1000 ?
El_Arrow
  • El_Arrow
A
SolomonZelman
  • SolomonZelman
A/100 OR A/1000 ? I don't understand you....
El_Arrow
  • El_Arrow
my bad A/100
SolomonZelman
  • SolomonZelman
Yes, because you are dividng by a smaller value, right?
El_Arrow
  • El_Arrow
yes
SolomonZelman
  • SolomonZelman
Now, which is greater 1/(n^2+1) OR 1/(n^2) ?
El_Arrow
  • El_Arrow
the second one
SolomonZelman
  • SolomonZelman
Yes, because you are dividng by a smaller value, thus obtaining a greater result.
El_Arrow
  • El_Arrow
so 1/(n^2) converges but 1/(n^2+1) diverges
El_Arrow
  • El_Arrow
i am using the comparison test and was wondering if that could happen? @SolomonZelman
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
  • 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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