Takeover Season
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Hi,

I was just wondering in my course, we have many tests, root test, ratio test etc. and they all are about sums from n = 1 to n = infinity etc. But, some questions such as proving if the exponential series is convergent, start from n = 0.

So, I was just wondering, if we have:
\sum _{n=1}^{\infty }\:a_n and it is convergent, then provided when n = 0, a_n = a_0 is finite, then can I also conclude that \sum _{n=0}^{\infty }\:a_n is convergent?

i.e. if I add on another term at n = 0 and it is finite, then it doesn't change convergence right?
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RDKGames
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(Original post by Takeover Season)
Hi,

I was just wondering in my course, we have many tests, root test, ratio test etc. and they all are about sums from n = 1 to n = infinity etc. But, some questions such as proving if the exponential series is convergent, start from n = 0.

So, I was just wondering, if we have:
\sum _{n=1}^{\infty }\:a_n and it is convergent, then provided when n = 0, a_n = a_0 is finite, then can I also conclude that \sum _{n=0}^{\infty }\:a_n is convergent?

i.e. if I add on another term at n = 0 and it is finite, then it doesn't change convergence right?
Of course!
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(Original post by RDKGames)
Of course!
Thank you!
So, if I had the question:
"Determine the values of x for which the series \displaystyle \sum _{n=0}^{\infty }\:\frac{x^n}{1+x^{2n}} converges."

Then, since when n = 0, the term in the summation is 1/2. I know this is finite, so I can just work with \displaystyle \sum _{n=1}^{\infty }\:\frac{x^n}{1+x^{2n}} instead right, because all my convergence/divergence tests use n = 1 to infinity sums etc.?

Also, do you have any tips on how to start off?
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RDKGames
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(Original post by Takeover Season)
Thank you!
So, if I had the question:
"Determine the values of x for which the series \displaystyle \sum _{n=0}^{\infty }\:\frac{x^n}{1+x^{2n}} converges."

Then, since when n = 0, the term in the summation is 1/2. I know this is finite, so I can just work with \displaystyle \sum _{n=1}^{\infty }\:\frac{x^n}{1+x^{2n}} instead right, because all my convergence/divergence tests use n = 1 to infinity sums etc.?

Also, do you have any tips on how to start off?
You can split it and think about it like that if you want. I just see that at the end of the day, we will have the same summand up to infinity so you need to use that summand in your convergence test.

I would use the root test here.

Notice that \displaystyle \lim_{n \to \infty} \dfrac{|x|}{\sqrt[n]{1+x^{2n}}} has two answers depending on whether |x|<1 or |x|>1.

Suppose |x| < 1 and proceed to determine a subset of this for which we have convergence via the root test.

Likewise with |x|>1.

You can check |x|=1 cases directly by plugging them in.
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(Original post by RDKGames)
You can split it and think about it like that if you want. I just see that at the end of the day, we will have the same summand up to infinity so you need to use that summand in your convergence test.

I would use the root test here.

Notice that \displaystyle \lim_{n \to \infty} \dfrac{|x|}{\sqrt[n]{1+x^{2n}}} has two answers depending on whether |x|<1 or |x|>1.

Suppose |x| < 1 and proceed to determine a subset of this for which we have convergence via the root test.

Likewise with |x|>1.

You can check |x|=1 cases directly by plugging them in.
Hi there,

Thank you for your reply. I really appreciate it. Yes, x = 1 gives us a constant function, x = -1 is quite clear standard ones we've seen - + - + etc..., I would usually use the Leibniz Alternating Series Theorem when I see (-1)^n but this case was quite simple, so I didn't need to.

Thank you, initially, I used the ratio test, i.e. I simplified \displaystyle \left(\frac{a_{n+1}}{a_n}\right) to get \displaystyle \frac{x\left(x^{2n}+1\right)}{1+  x^{2n+2}}. I was just wondering, can you give me tips on how can I find this limit any for what x it exists etc.
Not necessarily for what values of x is it less than 1, but rather how to actually calculate the limit for a general x, because if I can do that, I can do the rest e.g. check if it is < 1 etc...

But yes, I like your root test method, much simpler! Just wondering, you've taken \sqrt[n]{a}=\left|a\right|^{\frac{1}{n}}, i.e. when you take the positive square root, you just replace the thing inside with its absolute value?
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RDKGames
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(Original post by Takeover Season)
Hi there,

Thank you for your reply. I really appreciate it. Yes, x = 1 gives us a constant function, x = -1 is quite clear standard ones we've seen - + - + etc..., I would usually use the Leibniz Alternating Series Theorem when I see (-1)^n but this case was quite simple, so I didn't need to.

Thank you, initially, I used the ratio test, i.e. I simplified \displaystyle \left(\frac{a_{n+1}}{a_n}\right) to get \displaystyle \frac{x\left(x^{2n}+1\right)}{1+  x^{2n+2}}. I was just wondering, can you give me tips on how can I find this limit any for what x it exists etc.
Not necessarily for what values of x is it less than 1, but rather how to actually calculate the limit for a general x, because if I can do that, I can do the rest e.g. check if it is < 1 etc...

But yes, I like your root test method, much simpler! Just wondering, you've taken \sqrt[n]{a}=\left|a\right|^{\frac{1}{n}}, i.e. when you take the positive square root, you just replace the thing inside with its absolute value?
You can do the ratio test as well, although you will arrive at the same conclusion anyway. Don't forget that these tests use the absolute value in the limit.

\displaystyle L = \lim_{n \to \infty} \dfrac{|x|(1+x^{2n})}{1+x^{2n+2}  }

This depends on the value of x.

If |x| &lt; 1 then L = |x|. So L &lt; 1.

If |x| &gt; 1 then L = \dfrac{|x|}{x^2}. So L &lt; 1.

If |x| = 1, then L = 1. So the test is inconclusive.
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Takeover Season
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(Original post by RDKGames)
You can do the ratio test as well, although you will arrive at the same conclusion anyway. Don't forget that these tests use the absolute value in the limit.

\displaystyle L = \lim_{n \to \infty} \dfrac{|x|(1+x^{2n})}{1+x^{2n+2}  }

This depends on the value of x.

If |x| &lt; 1 then L = |x|. So L &lt; 1.

If |x| &gt; 1 then L = \dfrac{|x|}{x^2}. So L &lt; 1.

If |x| = 1, then L = 1. So the test is inconclusive.
Thanks for your reply dude. Just wondering, the way you calculated the limits.
I sort of did: \displaystyle \lim _{n\to \infty }\left(1+x^{2n+2}\right) = \lim _{n\to \infty }\left(1+\left(x^2\right)^{n+1}), so if for all |x| > 0, we have = \infty and for x = 0, we have = 1. Same argument for \displaystyle \lim _{n\to \infty }\left(1+x^{2n})=\lim _{n\to \infty }\left(1+\left(x^2\right)^n).

So:
\displaystyle L = \lim_{n \to \infty} \dfrac{|x|(1+x^{2n})}{1+x^{2n+2}  } = \left|x\right|\lim _{n\to \infty }\left(\frac{1+x^{2n}}{1+x^{2n+2  }}\right) and then I use L'Hopital's right for x \neq 0?

EDIT: May have confused myself with x and n, yeah, I get you sort of. Of course, \lim _{n\to \infty }\left(1+x^{2n}\right)=1+x^{\lim _{n\to \infty }\left(2n\right)}=1+x^{\infty } = \infty, if |x| &gt; 1

EDIT: Yes, okay, we can divide the fraction by x^{2n+2}.

Hi, I've got it lol, thank you so much!
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