KH_
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people please help xx

how do i tackle this question by applying ratio test? i know i should go by simplify An+1 /An but i could never reach the form
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RDKGames
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(Original post by KH_)
people please help xx

how do i tackle this question by applying ratio test? i know i should go by simplify An+1 /An but i could never reach the form
What question???
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KH_
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(Original post by RDKGames)
What question???
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sorry i wasnt able to add the picture .
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Zacken
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sorry i wasnt able to add the picture .
Like you said, the ratio test is a good shout here, you get:

\displaystyle \frac{5^{r+1} + 1}{(r+1)^{r+1} + 3} \times \frac{r^r + 3}{5^r + 1} < \frac{5^{r+1} + 5}{(r+1)^{r+1} + 3} \times \frac{r^r + 3}{5^r + 1} = \frac{5(r^r + 3)}{(r+1)^{r+1} + 3}

But \displaystyle \frac{5(r^r + 3)}{(r+1)^{r+1} + 3} < 5 \frac{r^r}{(r+1)^{r+1}} = 5\left(\frac{r}{r+1}\right)^r \times \frac{1}{r+1} = \cdots

then take the limit to show it goes to 0.
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KH_
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(Original post by Zacken)
Like you said, the ratio test is a good shout here, you get:

\displaystyle \frac{5^{r+1} + 1}{(r+1)^{r+1} + 3} \times \frac{r^r + 3}{5^r + 1} < \frac{5^{r+1} + 5}{(r+1)^{r+1} + 3} \times \frac{r^r + 3}{5^r + 1} = \frac{5(r^r + 3)}{(r+1)^{r+1} + 3}

But \displaystyle \frac{5(r^r + 3)}{(r+1)^{r+1} + 3} < 5 \frac{r^r}{(r+1)^{r+1}} = 5\left(\frac{r}{r+1}\right)^r \times \frac{1}{r+1} = \cdots

then take the limit to show it goes to 0.
thank you ! I wasn't thinking about using inequality ,it's quite helpful
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KH_
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(Original post by Zacken)
Like you said, the ratio test is a good shout here, you get:

\displaystyle \frac{5^{r+1} + 1}{(r+1)^{r+1} + 3} \times \frac{r^r + 3}{5^r + 1} < \frac{5^{r+1} + 5}{(r+1)^{r+1} + 3} \times \frac{r^r + 3}{5^r + 1} = \frac{5(r^r + 3)}{(r+1)^{r+1} + 3}

But \displaystyle \frac{5(r^r + 3)}{(r+1)^{r+1} + 3} < 5 \frac{r^r}{(r+1)^{r+1}} = 5\left(\frac{r}{r+1}\right)^r \times \frac{1}{r+1} = \cdots

then take the limit to show it goes to 0.

wait there's an error i think
shouldn't it be

[tex]\displaystyle \frac{5(r^r + 3)}{(r+1)^{r+1} + 3} > 5 \frac{r^r}{(r+1)^{r+1}}
and then the whole thing goes weird :/
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MadChickenMan
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(Original post by KH_)
wait there's an error i think
shouldn't it be

[tex]\displaystyle \frac{5(r^r + 3)}{(r+1)^{r+1} + 3} > 5 \frac{r^r}{(r+1)^{r+1}}
and then the whole thing goes weird :/
You're right. If you replace the constant with 20 say, instead of 5, that should work though.
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KH_
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Sorry , I don't quite understand , can you explain in detail ? The 5 thing is given by the question ?
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MadChickenMan
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(Original post by KH_)
Sorry , I don't quite understand , can you explain in detail ? The 5 thing is given by the question ?
I just mean that

\displaystyle \frac{5(r^{r}+3)}{(r+1)^{r+1}+3}<20\frac{r^{r}}{(r+1)^{r+1}}

In fact, for any constant greater than 5 the inequality will hold for r large enough. To hold for all r, the constant needs to be around 11.43, but of course any finite constant is sufficient to prove convergence.
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KH_
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(Original post by MadChickenMan)
I just mean that

\displaystyle \frac{5(r^{r}+3)}{(r+1)^{r+1}+3}<20\frac{r^{r}}{(r+1)^{r+1}}

In fact, for any constant greater than 5 the inequality will hold for r large enough. To hold for all r, the constant needs to be around 11.43, but of course any finite constant is sufficient to prove convergence.
Thanks for the explanation . But how do you get 11.43 ? I will have the diagnostic test in the fresher week ( which is inhumanite I know lol ) and I merely remember any Alevel maths and the more I revise the more worried I am
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MadChickenMan
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(Original post by KH_)
Thanks for the explanation . But how do you get 11.43 ? I will have the diagnostic test in the fresher week ( which is inhumanite I know lol ) and I merely remember any Alevel maths and the more I revise the more worried I am
If you rearrange the inequality you end with a function like this

\displaystyle \frac{1+\frac{3}{x^x}}{1+\frac{3}{(x+1)^{x+1}}}

which I graphed and observed that the maximum value was around 2.286 (5x2.286=11.43, hence why this is the required constant). You'd probably have a hard time differentiating the function and finding a zero.

What university is that at? Testing whether a series converges using things like the ratio test isn't usually taught until sometime in the first year, so surely that wouldn't come up on a diagnostic test?
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Zacken
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(Original post by KH_)
thank you ! I wasn't thinking about using inequality ,it's quite helpful
You're right, would be better to do something like this:

\displaystyle \frac{5(r^r + 3)}{(r+1)^{r+1} + 3} = \frac{5(1 + \frac{3}{r^r}}{\left(\frac{r+1}{r} \right)^r (r+1) + \frac{3}{r^r}} \to \frac{5(1 + 0)}{e(r+1) + 0}

so the denominator grows and the entire thing shoots to 0.
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