I still have problems understanding how to prove that a sequence converges, stuck with  $a_n=\left(\frac{n^2}{2^n} \right)$ I know that $\lim_{n\to\infty} a_n=0$ but don't know how to proceed. I tried replacing $2^n$ with $\sum_{k = 0}^n { \binom{n}{k}}$ to get $\left(\frac{n^2}{\sum_{k = 0}^n \left(\frac{n!}{k!(n-k)!} \right)} \right)$. Next I want to manipulate the term somehow (don't understand how this could be done) to get $n$ instead of $n^2$ as numerator to show that the denominator is always bigger than $n^2$ starting from some $n_0$. How would I then get the value of $\epsilon$?  Is this a valid approach? How can I extrace a $n$ from the sum? Also different approaches are appriciated!   @edit forgot to include that I want to show that this sequence converges to $0$, so showing convergence is not enough.  In the linked dublicate it's written as fact if $a_n &gt; 0$ and the ratio test $&lt; 1$. Why is this true?
38578
Prove that:  $$\lim_{n\rightarrow \infty}\dfrac {n^{2}}{2^{n}}=0$$  I think I need to show $2^n \geq n^3$ $\forall n \geq10$. Is it true?  What about: $$\lim_{n\rightarrow \infty}\dfrac {n^{2}}{n!}=0$$  What to do with $n!$?
88487
Let we have the following sequence $(a_n)$ such that $$a_n=\sum_{k=1}^n\frac{1}{n+k}$$ How can I prove that $(a_n)$ is increasing bounded sequence, then prove it is convergent and find its limit?
39151
I remember that a couple of years ago a friend showed me and some other people the following expression:  $$\lim_{n\to \infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}.$$  As shown below, I can prove that this limit exists by the monotone convergence theorem. I also remember that my friend gave a very dubious "proof" that the value of the limit is $\frac{1}{e-1}$. I cannot remember the details of the proof, but I am fairly certain that it made the common error of treating $n$ as a variable in some places at some times and as a constant in other places at other times. Nevertheless, numerical analysis suggests that the value my friend gave was correct, even if his methods were flawed. My question is then:  What is the value of this limit and how do we prove it rigorously?  (Also, for bonus points, What might my friend's original proof have been and what exactly was his error, if any?)  I give my convergence proof below in two parts. In both parts, I define the sequence $a_n$ by $a_n=\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}$ for all integers $n\ge 2$. First, I prove that $a_n$ is bounded above by $1$. Second, I prove that $a_n$ is increasing.   (1) The sequence $a_n$ satisfies $a_n&lt;1$ for all $n\ge 2$.  Note that $a_n&lt;1$ is equivalent to $1^n+2^n+\cdots+(n-1)^n&lt;n^n$. I prove this second statement by induction. Observe that $1^2=1&lt;4=2^2$. Now suppose that $1^n+2^n+\cdots+(n-1)^n&lt;n^n$ for some integer $n\ge 2$. Then  $$1^{n+1}+2^{n+1}+\cdots+(n-1)^{n+1}+n^{n+1}\le(n-1)(1^n+2^n+\cdots+(n-1)^n)+n^{n+1}&lt;(n-1)n^n+n^{n+1}&lt;(n+1)n^n+n^{n+1}\le n^{n+1}+(n+1)n^n+\binom{n+1}{2}n^{n-1}+\cdots+1=(n+1)^{n+1}.$$  (2) The sequence $a_n$ is increasing for all $n\ge 2$.  We must first prove the following preliminary proposition. (I'm not sure if "lemma" is appropriate for this.)  (2a) For all integers $n\ge 2$ and $2\le k\le n$, $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We observe that $k^2-1\le kn$, so upon division by $k(k^2-1)$, we get $\frac{1}{k}\le\frac{n}{k^2-1}$. By , we may find:  $$\frac{k+1}{k}\le 1+\frac{n}{k^2-1}\le\left(1+\frac{1}{k^2-1}\right)^n=\left(\frac{k^2}{k^2-1}\right)^n.$$  A little multiplication and we arrive at $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We may now first apply this to see that $\left(\frac{n-1}{n}\right)^n\le\left(\frac{n}{n+1}\right)^{n+1}$. Then we suppose that for some integer $2\le k\le n$, we have $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$. Then:  $$\left(\frac{k-1}{n}\right)^n=\left(\frac{k}{n}\right)^n\left(\frac{k-1}{k}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}\left(\frac{k}{k+1}\right)^{n+1}=\left(\frac{k}{n+1}\right)^{n+1}.$$  By backwards (finite) induction from $n$, we have that $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$ for all integers $1\le k\le n$, so:  $$a_n=\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots+\left(\frac{n-1}{n}\right)^n\le\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}&lt;\left(\frac{1}{n+1}\right)^{n+1}+\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}=a_{n+1}.$$  (In fact, this proves that $a_n$ is strictly increasing.) By the monotone convergence theorem, $a_n$ converges.  I should note that I am not especially well-practiced in proving these sorts of inequalities, so I may have given a significantly more complicated proof than necessary. If this is the case, feel free to explain in a comment or in your answer. I'd love to get a better grip on these inequalities in addition to finding out what the limit is.  Thanks!
98646
Show that the sequence $$\sqrt{2},\sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}},...$$ converges and find its limit. I put the sequence in this form , $(x_n)$ where $$\large x_n=2^{\Large\sum_{k=1}^{n}\left(\frac{1}{2^k}\right)}$$ I want to use Monotonic Convergence Theorem to show. But I stuck at proving the sequence is bounded above by 2. I manage to prove the sequence is an increasing sequence. Anyone can guide me ? I try to use induction to prove but i stuck at inductive step.
61842
Let $(a_n)$ be a sequence of positive terms and suppose that $\sum_\limits{n=0}^{\infty} a_n$ converges. Show that $\sum_\limits{n=0}^{\infty} a_n^2$ converges.   This is in the section on the Comparison Test so that must be what I'm supposed to use.  But I don't see how.  $(a_n)^2$ might be smaller or larger than $a_n$ depending on $a_n$.  And I can't use the Comparison Test with some other series because there's no info here about how fast $\sum a_n$ converges.  Hmm.  Any hints?
376
Let $a_n$ be a positive real sequence such that the series $\sum a_n$ converges.  I was asked to prove that under such circumstances $\sum a_n^{\frac{n}{n+1}}$ converges.  The previous sum can be rewritten as $\sum \frac{a_n}{a_n^{1/(n+1)}}$.  I can't prove the convergence of this last series... I'm sure comparison test is enough here, but other than $a_n \rightarrow 0$ there is no information that helps.  Thanks for your help.
84905
I need some verification for my proof in part a) and help to get me started on part b)  a) Prove that the sequence $a_n = (2n+1)/(3n+5)$ converges to $2/3$ directly from the definition of convergence of a sequence.  Solution: Some rough work first: Given $\epsilon&gt;0$, we want $|a_n-(2/3)|&lt; \epsilon$. Need $|(2n+1)/(3n+5) - (2/3)|&lt; \epsilon$ which after simplifying gives $7/(9n+15)&lt; \epsilon$. Solving for $n$ gives $n&gt; (7-15\epsilon)/9\epsilon$.  Then we write our formal proof: Given $\epsilon&gt;0$, set $N= (7-15\epsilon)/9\epsilon$. Let $n&gt;N$. Then $|a_n-(2/3)| = 7/(9n+15) &lt; 7/(9((7-15\epsilon)/9\epsilon)+15) =\epsilon$.  Thus by definition the sequence converges to 2/3.  b) Suppose that the sequence $\{a_n\}$ is not bounded above (that is, for any real number $x$, there is some $n$ so that $a_n&gt;x$). Prove that $\{a_n\}$ does not converge to $42$.  To solve this I am assuming that we have to suppose that $\{a_n\}$ does converge to $42$, and then we will get a contradiction, that is the sequence will be bounded above. So can anyone help me get started on this?
38305
Assume that $x_n$ is a sequence such that $x_n$ converge to a constant $x$. I would like to prove that   $\left( 1+ x_n/n \right)^n \rightarrow e^x$  Obvious if $x_n = x$ the result would be given. The extra sequence complicates things and I would like a hint or solution on how to proof this.
103850
I've looked up several related questions, but they do not answer what I am looking for.  Please give link if this is a duplicate.  What I eventually want to know is why  $$\lim_{n\to\infty}\left(1+\frac1n\right)^n$$  converges to a finite value.  I mean, we have to know that the limit converges before we can call it a constant, right?  I know that $f(n)=\left(1+\dfrac1n\right)^n$ is strictly increasing, but how do we know it is bounded above?  Also, I tried using the binomial theorem to write  $$\lim_{n\to\infty}\left(1+\frac1n\right)^n=\lim_{n\to\infty}\left[\sum_{k=0}^n\frac{n!}{n^k k! (n-k)!}\right],\tag{1}$$  but I am unsure how to prove that the sum converges for $n\to\infty,$ because the general term depends on both $n$ and $k.$    Here is what I think will work.  Let $a_{n,k}=\dfrac{n!}{n^k k! (n-k)!}.$  Then $a_{n+1,k+1}=\dfrac{(n+1)!}{(n+1)^{k+1} (k+1)! (n-k)!}.$  Using the Ratio Test, we get  $$\lim_{n\to\infty\atop k\to\infty}\left|\frac{a_{n+1,k+1}}{a_{n,k}}\right|=\lim_{n\to\infty\atop k\to\infty}\left[\frac{1}{k+1}\left(\frac{n}{n+1}\right)^k\right].$$  where the limit on the RHS is less than 1.    That may take care of the case where $n\in\mathbb{N},$ so I suppose that (1) can be generalized using "."
79337
Suppose $a_0$ is an arbitrary positive real number. Define the sequence $\{a_n\}$ by $$a_{n+1}=\frac{a_n+2}{a_n+1}$$ for all $n\geq0$. I have to prove that $\{a_n\}$ converges.  My attempt: If  $a=\lim_{n\to\infty}{a_n}$ exists, then it should be a solution to $$a=\frac{a+2}{a+1}$$ which is $\sqrt2$. Thus I need to show that $|\sqrt2 - a_n|$ gets arbitrarily small for large $n$. I tried to prove that $|\sqrt2-a_n|&lt;|\sqrt2-a_{n-1}|$ but couldn't.
71569
So one of my professors proposed a problem to me and it has stumped me for some time now. Here's how it goes:  Suppose you have a sequence $a_n$ of real numbers such that $$\lim_{n\to\infty} a_{n} = 0$$ and suppose the sequence of partial sums $s_n$ is bounded. Prove that $s_n$ converges or give a counterexample.  I'm hoping to figure this out without anyone handing me the complete solution, so if someone could point me in the right direction with a hint, it would be much appreciated.
119998
I need help proving $a_n=\left(\dfrac{n+1}{n}\right)^n$ is increasing sequence on the positive integers. An exercise in the analysis book by Mattuck asks to prove $a_n=\left(\dfrac{2^n+1}{2^n}\right)^{2^n}$ is increasing. But this is easy since $\left(\dfrac{2^n+1}{2^n}\right)^{2^n}=\left(\left(\dfrac{2^n+1}{2^n}\right)^{2}\right)^{2^{n-1}} =\left(\dfrac{2^{2n}+2\left(2^n\right)+1}{2^{2n}}\right)^{2^{n-1}}=$ $\left(\dfrac{2^{n-1}+1+\frac{1}{2^{n+1}}}{2^{n-1}}\right)^{2^{n-1}}&gt;\left(\dfrac{2^{n-1}+1}{2^{n-1}}\right)^{2^{n-1}}$ as desired. I am having difficulty in the general case: I must prove $\left(\frac{n+2}{n+1}\right)^{n+1}&gt;\left(\frac{n+1}{n}\right)^n$. This is equivalent to $(n+2)^{n+1}n^n&gt;(n+1)^{2n+1}$ which is the same as proving $\frac{n+2}{n+1}&gt;\left(\frac{(n+2)(n)}{(n+1)^2}\right)^n$ which is the same as $\frac{n+2}{n+1}&gt;\left(1-\frac{1}{(n+1)^2}\right)^n$ I'm stumped. Thanks in advance. Oh, and I don't want to use the derivative.
30158
Let $a_n$ be a sequence of nonnegative numbers. Then I want to show that      If $\,\sum_{n=1}^\infty a_n^2\,/ n\,$ converges, then $\frac 1N\!\sum_{n=1}^{N}a_n\,$ converges to zero.    A hint suggests me to write $a_n=\sqrt{n}\cdot \frac{a_n}{\sqrt n}$ and use the Cauchy-Schwarz inequality. However it does not seem enough. I cannot show that $\frac 1N\!\sum_{n=1}^{N}a_n$ converges to zero.  Could anyone please help me to prove it?
39500
Let $$ a_{n+1} =  \frac{1}{k} \left(a_{n} + \frac{k}{a_n}\right) ; k&gt;1, a_1&gt;0 $$ The problem is to show that it converges.  Attempt: The sequence is not monotone but it has a lower bound. It seems that odd terms subsequence and even term subsequence are monotonic sequences (I wrote some basic code to make this observation) though I am not able to prove it analytically. I also know that if odd subsequence and even subsequence converge to same limit then the sequence also converges. So, that tells me that I am on the right track.  Please provide any hints.
112504
I've already proved that for any $p &gt; 0$ and for any $\alpha \in \mathbb{R}$, the sequence $\frac{n^\alpha}{(1 + p)^n}$ converges to $0$.  Now, I want to prove that $\lim_{n \to \infty} x^n = 0$ as long as $|x| &lt; 1$.  I've split it into two possible cases:  (1) $0 &lt; x &lt; 1$.  Let $p = \frac{1}{x} - 1$, so that $x = \frac{1}{1 + p}$ and $p &gt; 0$ (since $x &lt; 1$). Then we let $\alpha = 0$ to see that $$\lim_{n \to \infty} x^n = \lim_{n \to \infty} \frac{n^0}{(1 + p)^n} = 0$$  (2) $-1 &lt; x &lt; 0$.  Here we cannot use the sequence $\frac{n^\alpha}{(1 + p)^n}$ again, since if we let $p = \frac{1}{x} - 1$, then $p$ isn't necessarily positive (for example, if $x = - \frac{1}{2}$, then $p = -3 &lt; 0$).  How can I prove Case (2)?
15022
Here's the original question:     Let $(a_n)$ be bounded. Assume that $a_{n+1} \ge a_{n} - 2^{-n}$. Show   that $(a_n)$ is convergent.   Okay, I know that if I can show that if the sequence is monotone, I can conclude that it is convergent. But I am not sure how to show that it is monotone.  I know that  $$a_n \le a_{n+1} + \frac{1}{2^n} &lt; a_{n+1} + \frac{1}{n}$$  It looks to me as if it is monotonically increasing but I'm quite not sure how to prove my claim. Any hints would be appreciated.
12869
I am playing around with the root/ratio test to practice with series. I just showed that $\sum \frac{1}{n!}$ converges by using the ratio test. I decided to see how things would go with the root test and I got stuck at something that I can't find on google. Right away while running the root test I encountered $\lim_{x \to +\infty}\left(\frac{1}{n!}\right)^\frac{1}{n}$. I have made the claim that $\left(n!\right)^\frac{1}{n}\geq 1 \hspace{3mm}\forall n\in \mathbb{N}.$ I have begun the proof and I think it is on the right track but some verification would be nice.  Proof: We can see that for n=1, this obviously holds, $\left(1!\right)^1\geq 1.$ Now suppose that this is the case for some $n=k$, then we have $\left((k+1)!\right)^{\frac{1}{k+1}}= \left((k+1)k!\right)^\frac{1}{k+1}$. It is at this point that I start to have a little trouble. Can somebody give me a push in the right direction? Thanks!