Forgot password?
 Register account
View 1879|Reply 9

[不等式] 放缩放缩放

[Copy link]

46

Threads

82

Posts

766

Credits

Credits
766

Show all posts

Shiki Posted 2019-7-21 22:40 |Read mode
试求$$\lceil \sum_{n=1}^{1000}\frac {1}{2n^{\frac{1}{3}}}\rceil$$和$$\lfloor \sum_{n=1}^{1000}\frac {1}{2n^{\frac{1}{3}}}\rfloor$$
= =

458

Threads

951

Posts

9832

Credits

Credits
9832

Show all posts

青青子衿 Posted 2019-7-21 23:17
回复 1# Shiki
\[ \operatorname{Floor}\left[\sum\limits_{n=1}^{1000}\dfrac{1}{\sqrt[3]{n}}\right]=149 \]
Floor[Sum[1/Power[n, 1/3], {n, 1000}]]

46

Threads

82

Posts

766

Credits

Credits
766

Show all posts

 Author| Shiki Posted 2019-7-21 23:37
回复 2# 青青子衿

感谢,请问是怎么解的?

686

Threads

110K

Posts

910K

Credits

Credits
91229
QQ

Show all posts

kuing Posted 2019-7-22 00:39
因为 `1/\sqrt[3]x` 递减且下凸,故
\[
\int_n^{n+1}\frac1{\sqrt[3]x}\rmd x
<\frac1{\sqrt[3]n}<
\int_{n-1/2}^{n+1/2}\frac1{\sqrt[3]x}\rmd x,
\]所以
\[
\frac12\int_1^{1000+1}\frac1{\sqrt[3]x}\rmd x
<\sum_{n=1}^{1000}\frac1{2\sqrt[3]n}<
\frac12\int_{1/2}^{1000+1/2}\frac1{\sqrt[3]x}\rmd x,
\]即
\[
\frac34(1001^{2/3}-1)
<\sum_{n=1}^{1000}\frac1{2\sqrt[3]n}<
\frac34\left( \left( 1000+\frac12 \right)^{2/3}-\left( \frac12 \right)^{2/3} \right),
\]显然 LHS `>3/4\cdot99=74.25` 以及
\[
\RHS<\frac34\left( 1000^{2/3}+\left( \frac12 \right)^{2/3}-\left( \frac12 \right)^{2/3} \right)=75,
\]所以原式在 `(74.25,75)` 内。

686

Threads

110K

Posts

910K

Credits

Credits
91229
QQ

Show all posts

kuing Posted 2019-7-22 00:54
上界其实也可以用 $<\displaystyle\int_{n-1}^n\frac1{\sqrt[3]x}\rmd x$ 来放缩,但由于 n=1 时涉及瑕积分,虽然其实没问题可以照积,但为免不必要的解释,还是避开它好了。
当然也可以保留第一项不放缩再用它(此时将得出 < 74.75),但我想让过程对称一点,所以利用凸来放。

46

Threads

82

Posts

766

Credits

Credits
766

Show all posts

 Author| Shiki Posted 2019-7-22 09:59
回复 5# kuing

感谢,稍后把参考答案未用积分放缩的方法贴上来

46

Threads

82

Posts

766

Credits

Credits
766

Show all posts

 Author| Shiki Posted 2019-7-22 12:25
本帖最后由 kuing 于 2019-7-22 14:27 编辑

手机写代码好累。。其实本质上就是积分放缩,只不过写成了装B形式...

注意到
\begin{align*}
\sum_{n=1}^{1000}\frac1{\sqrt[3]n}
&=\sum_{n=1}^{1000}\frac{n+1-n}{\sqrt[3]n}\\
&=\sum_{n=1}^{1000}\left(\frac{\sqrt[3]{(n+1)^2}-\sqrt[3]{n^2}}{\sqrt[3]n}\cdot
\frac{\sqrt[3]{n^2}+\sqrt[3]{(n+1)^2}+\sqrt[3]{n(n+1)}}{\sqrt[3]{n+1}+\sqrt[3]n}\right)\\
&>1+\sum_{n=2}^{999}\frac{3(\sqrt[3]{(n+1)^2}-\sqrt[3]{n^2})}2\\
&=1+\frac{3(100-\sqrt[3]4)}2 \\
&>1+\frac{3(100-2)}2=148,
\end{align*}

\begin{align*}
\sum_{n=1}^{1000}\frac1{\sqrt[3]n}
&=\sum_{n=1}^{1000}\frac{n-(n-1)}{\sqrt[3]n}\\
&=\sum_{n=1}^{1000}\left(\frac{\sqrt[3]{n^2}-\sqrt[3]{(n-1)^2}}{\sqrt[3]n}\cdot
\frac{\sqrt[3]{n^2}+\sqrt[3]{(n-1)^2}+\sqrt[3]{n(n-1)}}{\sqrt[3]{n-1}+\sqrt[3]n}\right)\\
&<1+\sum_{n=2}^{1000}\frac{3(\sqrt[3]{n^2}-\sqrt[3]{(n-1)^2})}2\\
&=\frac{299}2<150,
\end{align*}
于是
\[74<\sum_{n=1}^{1000}\frac1{2\sqrt[3]n}<75.\]

686

Threads

110K

Posts

910K

Credits

Credits
91229
QQ

Show all posts

kuing Posted 2019-7-22 13:25
回复 7# Shiki

手机写代码???
把过程拍个照发上来,我再帮你转码吧
如果发不了图,那就加我QQ发吧

46

Threads

82

Posts

766

Credits

Credits
766

Show all posts

 Author| Shiki Posted 2019-7-22 14:06
回复 8# kuing

已加qq

686

Threads

110K

Posts

910K

Credits

Credits
91229
QQ

Show all posts

kuing Posted 2019-7-22 14:29
回复 9# Shiki

OK,已将过程码至 7# 中。

Mobile version|Discuz Math Forum

2025-5-31 10:53 GMT+8

Powered by Discuz!

× Quick Reply To Top Edit