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maths diagnostic test

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

Reply 1

RDKGames

Study Forum Helper

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???

Reply 2

KH_

Original post by RDKGames

What question???

sorry i wasnt able to add the picture .

Reply 3

Zacken

Original post by KH_

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.

Reply 4

KH_

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 :smile:

Reply 5

KH_

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 :/

(edited 7 years ago)

Reply 6

MadChickenMan

Original post by KH_

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

\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.

Reply 7

KH_

Sorry , I don't quite understand , can you explain in detail ? The 5 thing is given by the question ?

Reply 8

MadChickenMan

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.

Reply 9

KH_

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}}

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 :frown:

Reply 10

MadChickenMan

Original post by KH_

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?