Test series for convergence

  • Thread starter DryRun
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  • #26
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I used the ratio test directly on the original series but it gets complicated and there is factor multiplication involved (as none of the terms cancel out) in both the numerator and denominator which expands so i'm not sure if it even resolves to some value.

Actually, you're right. I ended splitting it up at a later stage, probably where you got stuck.
 
  • #27
Bacle2
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Hope not to throw off the subject too much; just curious as to how testing convergence

of series became important to Calculus/Analysis. Is there something other than a

connection with Riemann sums?
 
  • #28
vela
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You need to clean up your notation.
[tex]\lim_{r\to \infty}\ln y=\frac{1/ \left( 1+\frac{1}{r} \right).(-1/r^2)}{-1/r^2}=\frac{1}{ \left( 1+\frac{1}{r} \right)}[/tex]
Should be
$$\lim_{r\to\infty}\ln y
= \lim_{r\to\infty} \frac{1/ \left( 1+\frac{1}{r} \right)\cdot(-1/r^2)}{-1/r^2}
= \lim_{r\to\infty} \frac{1}{1+\frac{1}{r}} = 1
$$ You need to include ##\displaystyle \lim_{r\to\infty}## until you actually evaluate the limit.

Same problem with
$$\ln \lim_{r\to \infty}y=\frac{1}{ \left( 1+\frac{1}{r} \right)}=\frac{r}{r+1}=1+\frac{1}{r} \\
\lim_{r\to \infty}y=e^{1+\frac{1}{r}}=e^1$$
To say ##\ln \displaystyle\lim_{r\to \infty}y## equals ##\frac{1}{ \left( 1+\frac{1}{r} \right)}## is wrong. The variable r won't appear in the result of the limit, yet what you've written says that the limit is equal to an expression that depends on r.

The sequence also strikes me as being a bit disjointed from what you wrote above. You found the limit of ln y, and then now you're talking about the log of the limit of y. Just for clarity, you might want to add the step
$$\ln \lim_{r\to\infty} y = \lim_{r\to\infty} \ln y = \cdots$$ so the grader can tell you know what you're doing. Alternatively, you could have also said
$$\lim_{r\to\infty} y = \lim_{r\to\infty} e^{\ln y} = \exp\left(\lim_{r\to\infty} \ln y\right) = e^1 = e$$
 
  • #29
DryRun
Gold Member
838
4
You need to clean up your notation.

Should be
$$\lim_{r\to\infty}\ln y
= \lim_{r\to\infty} \frac{1/ \left( 1+\frac{1}{r} \right)\cdot(-1/r^2)}{-1/r^2}
= \lim_{r\to\infty} \frac{1}{1+\frac{1}{r}} = 1
$$ You need to include ##\displaystyle \lim_{r\to\infty}## until you actually evaluate the limit.

Same problem with

To say ##\ln \displaystyle\lim_{r\to \infty}y## equals ##\frac{1}{ \left( 1+\frac{1}{r} \right)}## is wrong. The variable r won't appear in the result of the limit, yet what you've written says that the limit is equal to an expression that depends on r.

The sequence also strikes me as being a bit disjointed from what you wrote above. You found the limit of ln y, and then now you're talking about the log of the limit of y. Just for clarity, you might want to add the step
$$\ln \lim_{r\to\infty} y = \lim_{r\to\infty} \ln y = \cdots$$ so the grader can tell you know what you're doing. Alternatively, you could have also said
$$\lim_{r\to\infty} y = \lim_{r\to\infty} e^{\ln y} = \exp\left(\lim_{r\to\infty} \ln y\right) = e^1 = e$$

Thanks for the advice, vela. Actually, i have my finals coming up very soon, and i agree, my presentation of the limits isn't completely correct.
 

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