7

Question:

Edit: Original question is: Estimate $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{10000}$ to nearest hundred.

I used $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n}\geq \int_{0}^{n}\sqrt{x}dx$ and $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n}\leq \frac{4n+3}{6}\sqrt{n}$, which I need to prove.

So prove that: $$\left ( \frac{4n+3}{6} \right )\sqrt{n}+\sqrt{n+1}\leq \left ( \frac{4n+7}{6} \right )\sqrt{n+1}$$

Attempt: $$\left ( \frac{4n+3}{6} \right )\sqrt{n}+\sqrt{n+1}= \frac{(4n+3)\sqrt{n}+6\sqrt{n+1}}{6}\leq \frac{(4n+3)\sqrt{n+1}+6\sqrt{n+1}}{6}= \frac{(4n+9)\sqrt{n+1}}{6}$$

But $$\frac{(4n+9)\sqrt{n+1}}{6}\geq \frac{(4n+7)\sqrt{n+1}}{6}$$

How can I prove that $$\frac{(4n+3)\sqrt{n}+6\sqrt{n+1}}{6}\leq \frac{(4n+7)\sqrt{n+1}}{6}$$

please delete me
  • 1,201

1 Answers1

2

Your desired inequality

$$\frac{(4n+3)\sqrt{n}+6\sqrt{n+1}}{6}\leq \frac{(4n+7)\sqrt{n+1}}{6}$$

is equivalent to

$$(4n+3)\sqrt{n}\le(4n+1)\sqrt{n+1}\;.$$

Since we’re dealing with non-negative numbers here, this is equivalent to

$$n(16n^2+24n+9)\le(n+1)(16n^2+8n+1)\;.$$

If you expand that, you’ll see that it’s true.

Brian M. Scott
  • 631,399
  • Thank you very much! I noticed that when ever doing induction involving inequalities, it is best to remove any unnecessary terms. Simplifying the expression and leaving only what is necessary to prove makes it much easier. – please delete me Sep 03 '13 at 05:37
  • @user2357: This is why I asked do you really need to do induction. It is a straightforward non-induction manipulation. – André Nicolas Sep 03 '13 at 05:39
  • @user2357: You’re very welcome. – Brian M. Scott Sep 03 '13 at 05:39
  • @André: But this is the induction step of the OP’s argument. – Brian M. Scott Sep 03 '13 at 05:40
  • @AndréNicolas: $\left ( \frac{4n+3}{6} \right )\sqrt{n}+\sqrt{n+1}\leq \left ( \frac{4n+7}{6} \right )\sqrt{n+1}$ was the third step of induction because the original question involves a series. Is there another way without using induction? – please delete me Sep 03 '13 at 05:43
  • Sure, but one solution is enough. The error in the approximation is mostly due to the first few terms. So by calculating the first few terms "exactly" we can be more relaxed about error estimates for the rest. – André Nicolas Sep 03 '13 at 05:50