I'm trying to go through some questions on the ratio test at khan academy's integral calculus mission but I'm struggling a little.

​for the series $\dfrac{\sqrt{5x}}{(2x^2)+7}$

I've tried this: $\displaystyle\lim_{x\to\infty} \dfrac{(\sqrt{5x+1}) / (2x^{(2+1)}+7) }{ (\sqrt{5x}) / (2x^2+7)}$

And this: $\displaystyle\lim_{x\to\infty} \dfrac{(\sqrt{5x+5}) / (2x^2+2^2+7) } {(\sqrt{5x}) / (2x^2+7)}$

But neither gives an answer that is correct. According to the exercise the limit is $1$, but my attempts both give $0$.

I beleive the problem is my lack of understanding of how to properly apply $n+1$ for the ratio test and I'm having trouble finding information on its proper use. Could anyone help point me in the right direction, or towards anything I could read up on that might help?

Thanks in advance for your assistance! ​

  • $\begingroup$ Do you intend to examine the series $\sum_{n=0}^\infty\frac{\sqrt{5n}}{2n^2+7}$? $\endgroup$ – Lutz Lehmann Dec 1 '16 at 21:39
  • $\begingroup$ The second limit is clearly $1$. Note the double-fraction. $\endgroup$ – Peter Dec 1 '16 at 21:53

If you have

$$a_k = \frac{\sqrt{5k}}{2k^2 + 7}$$

then adding parenthesis around each $k$ makes it clearer what you are doing

$$a_k = \frac{\sqrt{5(k)}}{2(k)^2 + 7} $$

so when you need to find $a_{k+1}$, you write

$$a_{k+1} = \frac{\sqrt{5(k+1)}}{2(k+1)^2 + 7} = \frac{\sqrt{5k+5}}{2(k^2+2k+1) + 7} = \frac{\sqrt{5k+5}}{2k^2+4k+9}$$

EDIT: By the way, this is called substitution if you're looking for more of it to practice.

  • $\begingroup$ Thank you! That explains it perfectly. $\endgroup$ – Scottmeup Dec 4 '16 at 10:12

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