# Proofs 🥲 (1 Viewer)

#### Hello_World2

##### New Member
I don't have a clue of where to start this question? Can someone pls help ? Thanks !

#### Hello_World2

##### New Member
Bruh, for questions 4 and 5 I tried doing it, and then it got too complicated to solve. What am I doing wrong ??

Pls help thanks

#### Hello_World2

##### New Member
yess i got it thanks !

#### Drongoski

##### Well-Known Member
For Q5:

$\bg_white \text {Assume true for }n=k \geq 6 \implies \frac{1}{k!} < \frac {1}{e^k}\\ \\ \therefore \frac {1}{(k+1)!} = \frac {1}{k!} \times \frac {1}{k+1} < \frac {1}{e^k} \times \frac {1}{k+1} < \frac {1}{e^k} + \frac {1}{e} = \frac {1}{e^{k+1}} \\ \\ \text {since, for k greater than 5: } \frac {1}{k+1} <\frac {1}{e}$

So there you are.

Q4 requires quite a bit of LaTeX; so I'll give it a miss at the moment.

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#### hogzillaAnson

##### New Member
4) Let's skip straight to the inductive step.

\bg_white \begin{align*} \text{Assume that}\ \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{k} &< \sqrt{k} \\ \operatorname{LHS}_{k+1} = \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{k} + \frac{1}{k+1} &< \sqrt{k} + \frac{1}{k+1}\ \text{by assumption}\end{align*}
This is where it gets hard. We must show somehow that
$\bg_white \sqrt{k} + \frac{1}{k+1} < \sqrt{k+1}$
This is equivalent to proving that
$\bg_white \sqrt{k+1}-\sqrt{k} > \frac{1}{k+1}$
Consider that
\bg_white \begin{align*} \sqrt{k+1} - \sqrt{k} &= \frac{(\sqrt{k+1} - \sqrt{k})(\sqrt{k+1}+\sqrt{k})}{\sqrt{k+1} + \sqrt{k}} \\ &= \frac{1}{\sqrt{k+1}+\sqrt{k}} > \frac{1}{2\sqrt{k+1}} \\ &= \frac{\sqrt{k+1}}{2(k+1)} \geq \frac{\cancel{2}\sqrt{2}}{\cancel{2}(k+1)}\ \text{since}\ k \geq 7,\ \sqrt{k+1} \geq \sqrt{8}=2\sqrt{2} \\&> \frac{1}{k+1}\ \text{since}\ \sqrt{2} > 1\end{align*}
Thus, we can conclude that
$\bg_white \frac{1}{1} + \cdots + \frac{1}{k+1} < \sqrt{k+1}$
and by induction, the statement is true for all $\bg_white n \geq 7$

5) Considerably easier than 4 I think. Again, skip to the induction step

\bg_white \begin{align*} \text{Assume that}\ \frac{1}{k!} &< \frac{1}{e^k},\ \text{then} \\ \frac{1}{(k+1)!} &= \frac{1}{k+1} \cdot \frac{1}{k!} \\ &< \frac{1}{(k+1)e^k}\end{align*}
Now,
$\bg_white k+1 \geq 6+1 = 7 > e \implies \frac{1}{k+1} < \frac{1}{e}$
Therefore
$\bg_white \frac{1}{(k+1)!} < \frac{1}{(k+1)e^k} < \frac{1}{e^{k+1}}$
By P.O.M.I, the statement is true for all $\bg_white n\geq 6$

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