So far I have this. I have a feeling I'm getting off track with the last two steps.  We want to prove  $\sum_{i=1}^{n} \frac{1}{\sqrt{i}}$ $&lt; 2\sqrt{n}$ for $n \geq 1$  Base Case  Prove P(1): $\sum_{i=1}^{1} \frac{1}{\sqrt{1}}$ $&lt; 2\sqrt{1}$. We get $1 &lt; 2$.  Induction Hypothesis  $\sum_{i=1}^{n} \frac{1}{\sqrt{n}}$ $&lt; 2\sqrt{n}$ is true for $n \geq 1$  Induction Step  Prove P(k + 1): $\sum_{i=1}^{k + 1} \frac{1}{\sqrt{k + 1}} &lt; 2\sqrt{k + 1} $  LHS = $\sum_{i=1}^{k + 1} \frac{1}{\sqrt{k + 1}}$  = $\sum_{i=1}^{k} \frac{1}{\sqrt{k}}$ - 1 + $\frac{1}{\sqrt{k + 1}}$ + $\frac{1}{\sqrt{k + 2}}$  $&lt; 2\sqrt{k}$ - 1 + $\frac{1}{\sqrt{k + 1}}$ + $\frac{1}{\sqrt{k + 2}}$
33760
I am just starting out learning mathematical induction and I got this homework question to prove with induction but I am not managing. $$\sum_{k=1}^n\frac1{\sqrt k}&lt;2\sqrt n$$ Please help!
102016
As the title says I need to prove the following by induction: $$\sum_{i=1}^{2^n} \frac{1}{i} \ge 1+\frac{n}{2}, \forall n \in \mathbb N$$   When trying to prove that P(n+1) is true if P(n) is, then I get that:  $$\sum_{i=1}^{2^{n+1}} \frac{1}{i} = (\sum_{i=1}^{2^n} \frac{1}{i} ) +(\sum_{i=1}^{2^n} \frac{1}{2^n+i} ) \ge 1+\frac{n}{2} + (\sum_{i=1}^{2^n} \frac{1}{2^n+i} )$$ using the inductive hypothesis. And for P(n+1) to be true it would be sufficient to prove that  $$1+\frac{n}{2} + (\sum_{i=1}^{2^n} \frac{1}{2^n+i} ) \ge 1+\frac{n+1}{2}$$  or rewriting $$\sum_{i=1}^{2^n} \frac{1}{2^n+i} \ge \frac{1}{2} $$ and here I'm a bit stuck  Thank you for your help!
41759
I am just starting out learning mathematical induction and I got this homework question to prove with induction but I am not managing.  $$\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$$  Perhaps someone can help me out I don't understand how to move forward from here: $$\sum\limits_{k=1}^{n+1}{\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{n+1}}\ge \sqrt{n+1}}$$ proof and explanation would greatly be appreciated :) Thanks :)  EDIT sorry meant GE not = fixed :)
88141
So I have to prove:      For each natural number greater than or equal to 3,   $$(1+\frac1n)^n&lt;n$$   My work: Basis step: $n=3$ $$\left(1+\frac13\right)^3&lt;3$$ $$\left(\frac43\right)^3&lt;3$$ $$\left(\frac{64}{27}\right)&lt;3$$ which is true.  Now the inductive step, assume $P(k)=\left(1+\frac1k\right)^k&lt;k$ to be true and prove $P(k+1)=\left(1+\frac1{k+1}\right)^{k+1}&lt;k+1$.   This is where I am stuck because usually you add or multiply by $k+1$ or some similar term.
5998
Proving $\frac{1}{n+1} + \frac{1}{n+2}+\cdots+\frac{1}{2n} &gt; \frac{13}{24}$ for $n&gt;1,n\in\Bbb N$  To solve it I used induction but it is leading me nowhere my attempt was as follows:  Lets assume the inequality is true for $n = k$ then we need to prove that it is true for $k+1$  so we need to prove $\frac1{k+2} + \frac1{k+3}+\cdots+\frac1{2(k+1)} &gt; 13/24$  I don't know where to go from here please help.
25806
So for a better presentation:   I need to prove this using mathematical induction, so i started with a base of $n=1$  Since $i$ starts at $i = 1$ I got this:  $$\frac{1}{i(i+1)} = \frac{n}{n+1}$$ $$\frac{1}{2} = \frac{1}{2}$$  So since this was true I went to the induction part: I chose $n = k$ and assumed this was true:  $$\frac{1}{i(i+1)} = \frac{n}{n+1}$$ $$\frac{1}{i(i+1)} = \frac{k}{k+1}$$  but I got stuck here since I didn't know if I needed to increment $i$ or not and because I didn't know if I needed to add the summation symbol or not and needed to do a few things with it.
91788
Prove that for all $n \ge 1$:  $$\sum_{k=1}^n \frac{1}{k(k+1)} = \frac{n}{n+1}$$  What I have done currently:  Proved that theorem holds for the base case where n=1.  Then:  Assume that $P(n)$ is true. Now to prove that $P(n+1)$ is true:  $$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \frac{n}{n+1} + n+1$$  So: $$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \frac{n+1}{n+2}$$  However, how do I proceed from here?
66163
I ran across this problem in some old notes, and I frustratingly can't figure out how to do it   Let $a_i$ and $b_i$ be sequences of natural numbers, use induction to show  $\sum_{i=1}^n (a_ib_i)^{1/2} \le (\sum_{i=1}^n a_i)^{1/2}(\sum_{i=1}^n b_i)^{1/2} $  Obviously this is trivial to show for n=1. I can't make much progress on n+1. I've tried various tactics, squaring both sides etc.   Any hint or help would be appreciated. Thanks.
109392
Prove by induction $\sum \frac {1}{2^n} &lt; 1$  Well supposing the base case has been shown to be true, I start with the induction step:  Suppose true for n = k:  $$ \frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} &lt; 1$$  Want to show this is true for:  $$ \frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} + \frac{1}{2^{k+1}} &lt; 1$$  Now this is where i am getting stuck, should i be tryig to show that  $$ \frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} + \frac{1}{2^{k+1}} &lt; 1 +\frac{1}{2^{k+1}}$$ or should i be attempting to show that  $1 + \frac{1}{2^{k+1}} &lt; 1$ which is utter nonsense. So i am stuck on what exactly to prove here.  Note i also thought of maybe using the geometric series of $\frac {1}{2} $ as some upper bound but then that would provide me a value that was greater than 1 so it didn't work out
75476
Let $n$ be a positive natural number, $n\ge 2$. Then $\displaystyle\sum_{k=1}^n \frac{1}{k^2} \lt 2 - \frac{1}{n}.$   The basis step was easy but could someone give me a hint in the right direction as to how to do the induction step?  I tried this:  $\displaystyle\sum_{k=1}^k \frac{1}{k^2} + \frac{1}{(k + 1)^2} \lt 2 - \frac{1}{k + 1}$  But it's getting me nowhere or I am doing something wrong. I am no expert so a clear explanation would be appreciated. Thanks.
79185
Assumption: $\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}\ge\frac{2}{3}k\sqrt{k}$   Prove true for $n=k+1$ $$\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}+\sqrt{k+1}\ge\frac{2}{3}(k+1)\sqrt{k+1}$$  I'm upto :$$\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}+\sqrt{k+1}\ge\frac{2}{3}k\sqrt{k}+\sqrt{k+1}$$after which I'm stuck.
68740
This one should be pretty easy as its only a first year math problem. it s proof by induction that Im having trouble with.  $$let\;x_{1}=1; for\;each\;n\in \mathbb{N}\;let\;x_{n+1}=\frac{2}{3}x_{n}+1; prove\; x_{n}&lt; 3\;for\;all\;n$$  If anyone is willing to spell it out really obviously step by step and explain how they got from one step to the next I would appreciate it a lot. Thanks.
50713
Show that $\displaystyle\sum_{i=0}^l \dfrac{{l\choose i}}{{n\choose k + i}} = \dfrac{n+1}{(n-l+1) {n-l\choose k}}$ for all $k,l, n\geq 0$ with $k+l \leq n.$  The $k+l\leq n$ part doesn't seem that important. The claim clearly holds for $n=0,$ and in that case $k$ and $l$ must both be zero. I think the claim can be shown by induction, but I'm not sure how to do so; I'm not sure how to use the inductive hypothesis to prove the inductive step. The sum on the LHS is not always an integer, so finding a combinatorial argument doesn't seem very feasible. For $n = 1,$ we can have $(k,l) = (0,0), (1,0), (0,1),$ and in each case the sum holds. Perhaps multiplying both sides by a number (idk ${n-l\choose k}$) might be useful?
104983
$a _{1} = \frac{1}{2}, a _{2} = 1, a _{n}= \frac{1}{2} a_{n-1} + \sqrt{a _{n-2} }$  Show by the induction that it's $\le 4$ and also by the induction that $a_{n+1}-a_n \ge 0$  If it comes to first I don't know what to do and if it comes to second I tried many different ways but I couldn't prove it.
123132
I am having difficulty proving the following inequality: $$ \sqrt[n]{n} &lt; 1 + \sqrt{\frac{2}{n}} \quad \text{for all positive integers}\,\,\, n. $$ I am trying to use mathematical induction but I am having trouble going from the left side to the right side (in the induction step) and vice versa.   I also tried to take the power of $n$ on both sides and somehow use the binomial expansion but I am having difficulty with that. Thank you!
57076
Does anyone have any idea how to prove that  $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)...(n+k)} = \frac{1}{kk!}$$
49856
Question:     Let $P(n)$ be the statement that $1+\dfrac{1}{4}+\dfrac{1}{9}+\cdots +\dfrac{1}{n^2} &lt;2- \dfrac{1}{n}$. Prove by mathematical induction. Use $P(2)$ for base case.   Attempt at solution:  So I plugged in $P(2)$ for the base case, providing me with $\dfrac{1}{4} &lt; \dfrac{3}{2}$ , which is true.   I assume $P(n)$ is true, so I need to prove $P(k) \implies P(k+1)$.   So $\dfrac{1}{(k+1)^2} &lt; 2 - \dfrac{1}{k+1}$.  I don't know where to go from here, do I assume that by the Inductive hypothesis that it's true?