The server has encountered a problem because the resource was not found.
Your request was :
http://www.cs.berkeley.edu/~gvaliant/papers/KMV_full.pdf
What are you looking for?Maxima在微积分上之应用无限的数列_百度文库
两大类热门资源免费畅读
续费一年阅读会员，立省24元！
评价文档：
38页免费38页免费4页1下载券2页¥1.003页¥1.00 2页1下载券2页1下载券5页1下载券24页1下载券23页免费
喜欢此文档的还喜欢35页免费17页免费95页免费24页免费57页免费
Maxima在微积分上之应用无限的数列|M​a​x​i​m​a​在​微​积​分​上​之​应​用​无​限​的​数​列
把文档贴到Blog、BBS或个人站等：
普通尺寸(450*500pix)
较大尺寸(630*500pix)
你可能喜欢Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
To prove the convergence of
$$\sum_{n=1}^{\infty} \frac1{n^p}$$
for $p & 1$, one typically appeals to either the Integral Test or the Cauchy Condensation Test.
I am wondering if there is a self-contained proof that this series converges which does not rely on either test.
I suspect that any proof would have to use the ideas behind one of these two tests.
45.2k586205
We can bound the partial sums by multiples of themselves:
$$\begin{eqnarray}
\sum_{n=1}^{2k+1}\frac{1}{n^p}\\
1+\sum_{i=1}^k\left(\frac{1}{(2i)^p}+\frac{1}{(2i+1)^p}\right)\\
&&&1+\sum_{i=1}^k\frac{2}{(2i)^p}\\
&=&1+2^{1-p}S_k\\
&&&1+2^{1-p}S_{2k+1}\;.
\end{eqnarray}$$
Then solving for $S_{2k+1}$ yields
$$S_{2k+1}&\frac{1}{1-2^{1-p}}\;,$$
and since the sequence of partial sums is monotonically increasing and bounded from above, it converges.
102k8106209
My personal favourite is a variant of a common proof that the harmonic series diverges: we get
$$\sum_{n=2^k}^{2^{k+1}-1}\frac1{n^p}\le2^k\cdot\frac1{2^{kp}}=2^{(1-p)k}.$$ because the sum has $2^k$ terms of which the first is the largest. Now sum this over $k$ to get $$\sum_{n=1}^{\infty}\frac1{n^p}\le\sum_{k=0}^\infty2^{(1-p)k}=\frac1{1-2^{1-p}}&\infty$$ since $2^{1-p}&1$.
20.5k12238
FWIW, the properties of the generalized harmonic series $\sum \frac{1}{n^p}$ can be used to prove another convergence criterion (which goes under the name of second Cauchy's convergence criterion, as far as I remember).
The criterion is the following:
Let $(a_n)$ be a sequence of positive numbers. If:
$$\lim_{n\to \infty} \frac{\ln \frac{1}{a_n}}{\ln n} =L&1 \tag{1}$$
then the series $\sum a_n$ converges. On the other hand, if:
$$\lim_{n\to \infty} \frac{\ln \frac{1}{a_n}}{\ln n} =l&1 \tag{2}$$
then the series $\sum a_n$ diverges.
The proof is very simple. The criterion remains valid even if one replaces $\displaystyle \lim_{n\to \infty}$ with $\displaystyle \liminf_{n\to \infty}$ in (1) and with $\displaystyle \limsup_{n\to \infty}$ in (2).
This is the most direct and elementary way I know how to prove the result, although it only works for powers in the range $[0,1] \cup [2,\infty)$ which is exactly the uninteresting set, and Joriki's answer is much better regardless.
I had already written this, and perhaps somebody finds it useful.
First we consider $p=1$.
Set $x_n$ equal to the greatest power of 2 less than $\frac{1}{n}$. That is, $$(x_n) = (\frac{1}{2}, \frac{1}{4},\frac{1}{4}, \frac{1}{8}, \frac{1}{8},\frac{1}{8},\frac{1}{8},\dots).$$
$$\sum_{i=1}^{\infty} x_n
= \sum_{i=0}^{\infty}\Big(\sum_{j=2^i}^{2^{i+1}-1}x_j\Big)
= \sum_{i=0}^{\infty}\Big(\sum_{j=2^i}^{2^{i+1}-1} \frac{1}{2^{i+1}}\Big)
= \sum_{i=1}^{\infty} \frac{1}{2},
which diverges, and that $x_n & \frac{1}{n}$, which proves $\sum_{n=1}^{\infty} \frac{1}{n}$ also diverges.
If $p = 2$, let $x_n = \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$ if $n&1$ and 1 if $n=1$.
Observe that
$$\sum_{n=1}^{\infty} x_n = 1 + \sum_{n=2}^{\infty}\Big(\frac{1}{n-1} - \frac{1}{n}\Big) = 1 + 1 - \lim_{n\rightarrow\infty}(1/n) = 2,$$
and so in particular this series converges.
$x_n & \frac{1}{n \cdot n} = \frac{1}{n^2},$
so the series $$\sum_{n=1}^{\infty} \frac{1}{n^2}$$ converges as well.
Finally if $p&2$ then since $\frac{1}{n^p} & \frac{1}{n^2}$, the series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges, and
if $0&p&1$ we have $\frac{1}{n^p} & \frac{1}{n}$, and so $\sum_{n=1}^{\infty} \frac{1}{n^p}$ diverges.
For every $p&1$, one can reduce the problem to the convergence of some geometric series. Then, either one takes the latter for granted or one proves it by an elementary argument. More details below.
Let $N(k)$ denote the integer part of $a^k$, where the real number $a&1$ is defined by $a^{p-1}=2$. The sum of $1/n^p$ over $n$ between $N(k)$ and $N(k+1)$ is at most the number of terms $N(k+1)-N(k)$ times the greatest term $1/N(k)^p$. This contribution behaves like
(a^{k+1}-a^k)/a^{kp}\sim(a-1)a^{k(1-p)}=(a-1)2^{-k}.
The geometric series of general term $2^{-k}$ converges hence the original series converges.
There is a variant of this, where one considers the slabs of integers limited by the powers of $2$. The contribution of slab $k$ is then at most the number of terms $2^k$ times the greatest term $1/2^{kp}$. This reduces the problem to the convergence of the geometric series of general term $b^k$, where $b=2^{1-p}&1$.
Finally, as everybody knows, a simple proof of the convergence of the geometric series of general term $b^k$, for every $b$ in $(0,1)$ say, is to note that the partial sums $s_k$ are such that $s_0=1$ and $s_{k+1}=1+bs_k$, and to show by induction on $k$ that $s_k\le1/(1-b)$ for every $k\ge0$.
Edit The so-called variant above shows, using $b=2^{1-p}$, that the sum of the series is at most $1/(1-b)=2^p/(2^p-2)$. But the contribution of slab $k$ is also at least the number of terms $2^k$ times the smallest term $1/2^{(k+1)p}$, hence the sum of the series is at least $1/(2^p-2)$. Finally the sum of the series is between $1/(2^p-2)$ and $2^p/(2^p-2)$.
195k17135306
I suppose what joriki proved could be phrased as $$|\zeta(s)| \leq \frac{1}{1-2^{1-\sigma}} \ \ \ \ \ \forall \sigma &1 , \ \ s=\sigma+it$$ Instead of taking the sum in his second line over $\mathbb{Z}_2,$ one could do this over $\mathbb{Z}_q$ for any integer $q&1$ to get the nice inequality $$\sum_{n & q}\frac{1}{n^\sigma} \leq q^{1-\sigma} \ \ \zeta(\sigma) \ \ \ \ \ \ \ \ \forall \sigma&1, \ q \geq 2, q \in \mathbb{Z}$$ Now given any real $x&2$ we can use $q=1+[x]$ in the case that $x$ is not an integer to get
$$\zeta_{x}(\sigma):=\sum_{n & x}\frac{1}{n^\sigma} \leq x^{1-\sigma} \ \ \zeta(\sigma) +O(\frac{1}{x^{\sigma}})\ \ \ \ \ \ \ \ \forall \sigma&1, \ x \geq 2, x \in \mathbb{R}$$ and we can therefore phrase this as $$\zeta(s)=\zeta_{x}(s)+O(\frac{x^{1-\sigma}}{1-\sigma}+\frac{1}{x^{\sigma}}) \ \ \ \ \forall \sigma&1, \ x \geq 2, x \in \mathbb{R}$$ uniformly in the stated regions. This inequality might be of some interest regarding the truncated zeta function.
Here is another proof:
By Bernoulli
$$\frac{(n+1)^p}{n^p} \geq 1+ \frac{p}{n}=\frac{n+1}{n}+\frac{p-1}{n}$$
Multiplying by $n$ and dividing by $(n+1)^{p}$ you get
$$\frac{1}{n^{p-1}}\geq \frac{1}{(n+1)^{p-1}} + \frac{p-1}{(n+1)^p}$$
$$\frac{p-1}{(n+1)^p} \leq \frac{1}{n^{p-1}}-\frac{1}{(n+1)^{p-1}}
$$(p-1) \sum_{n=2}^{m+1} \frac{1}{n^p}=(p-1) \sum_{n=1}^m \frac{1}{(n+1)^p} \leq
\sum_{n=1}^m
\frac{1}{n^{p-1}}-\frac{1}{(n+1)^{p-1}}\,.$$
Since the last sum is telescopic you get
$$(p-1) \sum_{n=2}^{m+1} \frac{1}{n^p} \leq 1- \frac{1}{(m+1)^{p-1}}$$
53.3k253113
Let $S(n)$ = $ \Sigma_{1}^{n} \frac{1}{n^p}$
$S(2n)$ = $S(n)$ + $\frac{1}{{(n+1)}^p} $ +
$\frac{1}{{(n+2)}^p} $ + $\frac{1}{{(n+3)}^p} $ +$\frac{1}{{(n+4)}^p} $ ...........+ $\frac{1}{{(n+n)}^p} $
Let $\Delta S$ = $S(2n) - S(n)$
$\frac{n}{{(2n)}^p}$
$\leq$ $\Delta S$ $\leq$
$\frac{n}{(n+1)^p} $
By sandwitch theorem we see that if $p & 1$,
$\lim_{n \rightarrow \inf}$ $\Delta S = 0$
so that as n tends to infinity
$S(n) = S(2n) = S(4n) ....$
So that series converges
$S(2n) - S(n) \leq
\frac{n}{(n+1)^p}
\leq n^{1-p} $
$S(2^{k+1}n) - S(2^{k}n) \leq
\frac{2^{k+1}n - 2^{k}n}{(2^kn+1)^p} \leq \frac{2^{k+1}n - 2^{k}n}{(2^kn)^p} \leq n^{1-p}2^{k(1-p)}$
Summing above equation for k = 0 to m, we get
$S(2^{m+1}n) - S(n) \leq
n^{1-p}(\frac{1-2^{(m+1)(1-p)}}{1-2^{(1-p)}})$
Hence $\lim_{n \to \infty} R(n) =\lim_{m \to \infty}S(2^{m+1}n) - S(n) =0 $
Your Answer
Sign up or
Sign up using Google
Sign up using Facebook
Sign up using Stack Exchange
Post as a guest
required, but not shown
Post as a guest
required, but not shown
By posting your answer, you agree to the
Not the answer you're looking for?
Browse other questions tagged
Upcoming Events
Top questions and answers
Important announcements
Unanswered questions
By subscribing, you agree to the
Mathematics Stack Exchange works best with JavaScript enabledERY-LIKE FORMULAE FOR i(4n + 3)_百度文库
两大类热门资源免费畅读
续费一年阅读会员，立省24元！
评价文档：
15页免费32页免费23页免费7页免费5页免费 10页免费8页免费4页免费4页免费29页1下载券
喜欢此文档的还喜欢163页免费206页免费6页1下载券2页1下载券21页3下载券
ERY-LIKE FORMULAE FOR i(4n + 3)|T​h​e​ ​R​i​e​m​a​n​n​ ​z​e​t​a​-​f​u​n​c​t​i​o​n​ ​i​s​ ​d​e​f​i​n​e​d​ ​b​y​ ​i​(​s​)​:​=
把文档贴到Blog、BBS或个人站等：
普通尺寸(450*500pix)
较大尺寸(630*500pix)
你可能喜欢}