0

I'm trying to prove the following proposition by induction for all $n \in \mathbb{N} $ and $n \geq 2 $

$$ \sum_{j=1}^{k} \frac{1}{\sqrt{j}} \gt \sqrt{k} $$

However, I haven't been able to find a way to insert the induction hypothesis into the induction step to prove that this inequality holds for $n = k +1$.

A detailed, step-by-step explanation would be very much appreciated.

lone student
  • 14,709
dpakasa
  • 125

3 Answers3

4

You don't need induction to show that. Just observe that $$\sum_{j=1}^{k} \frac{1}{\sqrt{j}} \gt\sum_{j=1}^k\frac{1}{\sqrt k}= \sqrt{k}$$

Martund
  • 14,706
  • 2
  • 13
  • 30
  • 4
    Please read https://math.meta.stackexchange.com/questions/25438/questions-that-specify-an-approach. – player3236 Apr 10 '21 at 14:34
  • @player3236, the top answer to the post you linked justifies my answer. There are already some answers that use induction, I am just showing that induction is not at all needed to solve a problem as simple as this. Additionally, the original question in this case is not meant to learn induction, but to prove this hypothesis. I believe, my answer do not violate the guidelines given in the post you linked. – Martund Apr 10 '21 at 14:53
  • 2
    The "top" answer with 0 net upvotes, or the top answer with 25 upvotes posted by a diamond moderator? In any case, your answer is the only one that does not utilize induction when the question specifically asked for a proof using it - in other words, this is not an answer. – player3236 Apr 10 '21 at 15:18
  • @player3236, Any sensible definition of "top" can't call an answer with 0 net upvotes to be "top", so I was pretty clear what I mean by that. If you still don't think my answer makes sense, read the top answer again. – Martund Apr 10 '21 at 15:24
1

This is a consequence of

$$\begin{aligned} \sqrt{k+1} - \sqrt k &= (\sqrt{k+1} - \sqrt k)\frac{\sqrt{k+1} + \sqrt k}{\sqrt{k+1} + \sqrt k}\\ &= \frac{1}{\sqrt{k+1} + \sqrt k}\\ &\lt \frac{1}{\sqrt{k+1}} \end{aligned}$$

that you can use to move by induction from the hypothesis valid for $k$ to $k+1$:

if

$$ \sum_{j=1}^{k} \frac{1}{\sqrt{j}} \gt \sqrt{k} $$ then

$$ \sum_{j=1}^{k+1} \frac{1}{\sqrt{j}} = \sum_{j=1}^{k} \frac{1}{\sqrt{j}} + \frac{1}{\sqrt{k+1}} \gt \sqrt k + \sqrt{k+1} - \sqrt k = \sqrt{k+1}$$

mathcounterexamples.net
  • 70,018
1

Task: Prove the inequality using induction

$$\sum_{j=1}^{n} \frac{1}{\sqrt{j}} \gt \sqrt{n}$$

For $n=2$, then the statement is correct.

Suppose that, for $n=k$ the statement is also correct.

$$1+\frac {1}{\sqrt 2}+\frac {1}{\sqrt 3}+\cdots +\frac {1}{\sqrt k}>\sqrt k$$

Then, by induction for $k+1$, we get

$$\frac {1}{\sqrt 2}+\frac {1}{\sqrt 3}+\cdots +\frac {1}{\sqrt {k+1}}>\sqrt k+\frac {1}{\sqrt {k+1}}$$

Finally, it is enough to show that

$$\sqrt k+\frac {1}{\sqrt {k+1}}>\sqrt {k+1}$$

Thus we have,

$$\sqrt k+\frac {1}{\sqrt {k+1}}>\sqrt {k+1}$$

$$\frac {1}{\sqrt {k+1}}>\sqrt {k+1}-\sqrt k$$

$$\begin{align}\sqrt {k+1}&<\frac{1}{\sqrt {k+1}-\sqrt k}\\ &=\sqrt k+\sqrt{k+1}.\end{align}$$

which is correct.

lone student
  • 14,709