The inequality is  $1000n^3&lt;2^n$ I'm trying to see for what value of $n$ an algorithm A who takes ($1000n^3$) steps takes less time than another algorithm B who takes($2^n$) steps.  Any idea how to do that step by step?
121765
I'm new to this website. I've checked the other topics and couldn't find the answer to my question. I've also tried to find the result with TI-CAS but the only thing it does is simplify the inequality which really doesn't help at all.  the equation is : $1000n^3 &lt; 2^n$ Now I've been trying to use log but it gets me nowhere (I haven't seen this form before), the answer is supposed to be 24 but I'm not sure. So my question is how to solve step by step this inequality for n. Could you guys help me with this one?
14809
Let $n$ and $1\leq k \leq n$ be natural numbers. Prove the inequality     $$\sum_{i=k}^n \binom{n}{i}\bigg(\frac{k}{n+1}\bigg)^i\bigg(1-\frac{k}{n+1}\bigg)^{n-i} \leq 1 - \frac{1}{e} $$  Equivalently, if $X\sim$ Bin($n$,$\frac{k}{n+1}$), prove that $\mathbb{P}[X\geq k] \leq 1 - \frac{1}{e}$.   My attempt: It may be helpful to show that the LHS tends to $1-\frac{1}{e}$ as $n \to \infty$ (already did that) and that the LHS is an increasing function on $n$ (have not done that).
88419
How to solve this summation ? Also, I'm not sure what does $1\leq l&lt; m &lt;n$ supposed to imply in the development of summation form.  $$\sum_{1\leq l&lt; m &lt;n}^{} \frac{1}{5^l3^m2^n}$$  This one is from the Galois-Noether Contest in 2018:
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These are all from Sipser's book, second edition. I was just hoping someone could verify/explain those that are more difficult for me.  $2n = O(n)$: true  $n^2 = O(n)$: false  $n^2 = O(n\log^2 n)$: I say false, because $\log^2 n$ grows more slowly than $n$  $n\log n = O(n^2$): true because $n$ grows more quickly than $\log^2n$  $3^n = 2^{O(n)}$: I wasn't sure what this notation meant. Any clarification here?  $2^{2^n} = O(2^{2^n})$: I would say true, but I'm not entirely sure why.  $n = o(2n)$: false, because $o(2n) = o(n)$ and $n$ grows as quickly as itself.  $2n = o(n^2)$: true, $n^2$ is always more than $2n$ for some constant.  $2^n = o(3^n)$: true  $1 = o(n)$: I'm not really sure.  $n = o(\log n)$: false, $n$ is always larger.  $1 = o(1/n)$: true, because 1 is always larger.
82348
Prove that $\tau(2^n-1) \geq \tau(n)$ for all positive integers $n$.  Note that $\tau(2^n-1)=\sum_{d|2^n-1}{1}$. I try to prove by induction.  Base case: When $n=1$, we have $\tau(2-1)=\tau(1)$. Hence, the inequality is true.  Fix an $n$. Assume that the inequality $\tau(2^n-1) \geq \tau(n)$, which is $\sum_{d|2^n-1}{1} \geq \sum_{d|n}{1}$ is true for $n$. Note that $\sum_{d|2^{n+1}-1}{1}=\sum_{d|2^{n}-1}{1}+1 \geq \sum_{d|n}{1}+1=\sum_{d|n+1}{1}$  I don know whether the equality above hold or not. Can anyone guide me?
111624
I was trying to prove following inequality:  $$|\sin n\theta| \leq n\sin \theta \ \text{for all n=1,2,3... and } \  0&lt;\theta&lt;π $$  I succeeded in proving this via induction but I didn't get "feel" over the proof. Are there other proof for this inequality?
8473
Problem  For what natural numbers is $n^3 &lt; 2^n$?  Attempt @ Solution   For $n=1$, $1 &lt; 2$ Suppose $n^3 &lt; 2^n$ for some $n = k \ge 1$ It looks like the inequality is true for $n = 0$, $n = 1$ and $n\ge10$ But, how can I prove this through induction?
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I would appreciate if somebody could help me with the following problem:      For a given positive integer $n$, can we find positive integers $a$ and $k$ ($k\geq 2$) such that $2^n-1=a^k$?    The answer would be no, I  guess.  Please give me some ideas!
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How can I prove this inequality for all natural numbers $n\geq 2$?  ${2n\choose n}&gt;\frac{4^n}{n+1}$  I've tried induction but that was a dead end.
33562
I can show that for $x &gt; 0$ and $r_{i} &gt; 0$ we have $$ \left(\, x + r_{1}\,\right)\ldots\left(\, x + r_{n}\,\right)\ \geq\ \left[\, x + \left(\, r_{1}\ldots r_{n}\,\right)^{1/n}\,\right]^{n}.$$ However, I can't do this using straight up induction, strong or weak. Can someone do this?   You can show the inequality holds for $n$ if it holds for $n+1$ using $r_{n+1} = (r_1 \cdots r_n)^{1/n}$, so going down isn't a problem, but going up has eluded me. Additionally, you can show that if it holds for $n=a$ and $n=b$ then it holds for $n=ab$. Using powers of two and the above is sort of a way to prove the above by induction but it certainly isn't "normal".  You can also prove the above using Lagrange multipliers (which isn't surprising). The last proof I know is where you compare coefficients of $x^k$ and use the AM-GM inequality.
88351
I would like to prove the following inequality, for $n=0,1,2,...$,   $$ \binom{2n}{n}\leq 4^n.$$ I already proved it by induction, and I'm looking for another proof.
44829
So I know that showing $\left(1+\frac{1}{n}\right)^n$ is an increasing sequence has probably appeared on this site about 100 times, but my professor said he thinks the induction step is most easily seen if you expand this out and compare the summation term by term. I tried this and I did not what see why the inequality was obvious. Can somebody help me out here? Thanks!
97058
Comparing $\;2^n\;$ and $\;n!\;$  I need to rewrite one or the other so they can be more easily compared (i.e. have similar form). Because of the factorial, I'm a little lost as to how to compare the two functions. Normally, I would take the logarithm when trying to re-express a power like $2^n$, but I don't think this will work, since I don't know how to take the logarithm of a factorial, if at all.   Might the series expansions of either functions be the right approach? If not could, someone point me in the right direction?
3540
I don't know what to do with this. Nothing works. I hope somebody can help me to find a decision $$n!\leq {\left(\frac{n+1}{2}\right)}^{n}$$
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I'm given a inequality as such: $n! &gt; n^3$ Where n > 5,   I've done this so far:  BC: n = 6, 6! > 720 (Works)  IH: let n = k, we have that: $k! &gt; k^3$  IS: try n = k+1, (I'm told to only work from one side)  So I have (k+1)!, but I'm not sure where to go from here.  I've been told that writing out: $(k+1)! &gt; (k+1)^3$ is a fallacy, because I can't sub in k+1 into both sides, but rather prove from one side only.    Any Help on how to continue from here, would be much appreciated.
116601
If $a$ and $b$ are non-negative real numbers then show that $$(3a +\frac{4}{a+1}+\frac{8}{\sqrt{2(1+b^2)}})\cdot(3b +\frac{4}{b+1}+\frac{8}{\sqrt{2(1+a^2)}})\geq81.$$ Inequality is less common. To demonstrate the usual instruments (means inequality, ...) did not help. Does anyone have any idea?
43898
I am currently in the process of learning how to prove statements by induction. For the most part, I understand but I am at most clueless when it comes to prove some inequalities.  For example, I am having a hard time proving that: $n &lt; 2^n$ is true. (Even if it's clear that it is indeed true!)  In the induction step, I am stuck with: $n + 1 &lt; 2^{n + 1}$  I don't know what would be the assumption I should make from this statement.  Anyone could point me in the right direction? Thanks!