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# maths diagnostic test watch

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
2. (Original post by KH_)

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???
3. (Original post by RDKGames)
What question???

sorry i wasnt able to add the picture .
4. (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:

But

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

But

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

But

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 :/
7. (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.
8. Sorry , I don't quite understand , can you explain in detail ? The 5 thing is given by the question ?
9. (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

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.
I just mean that

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
11. (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

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?
12. (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:

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

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