Sheihuimin, Yun, Yi-Huai, Chanding-bian compilation of mathematical Analysis exercises Lecture notes section 16.2 exercises reference answers [from Tao younger brother]

1. Set the known $ \sum\limits_{n = 1}^\infty {{{\left ({-1} \right)}^{n-1}}{a_n}} = A,\sum\limits_{n = 1}^\infty {{a_{2n-1 }}} = B $, proof: $ \sum\limits_{n = 1}^\infty {{A_n}} $ converge and ask for and.
Solution: There is obviously
\[\sum\limits_{n = 1}^\infty {{A_n}} =2\sum\limits_{n = 1}^\infty {{a_{2n-1}}}-\sum\limits_{n = 1}^\infty{{{\left ( {-1} \right)}^{n-1}}{a_n}} = 2b-a.\]

2. Set $ P (x) =a_0+a_1x+\cdots+a_mx^m$ to $ M $ polynomial, order $ \sum\limits_{n= 0}^\infty {\frac{{p\left (n \right)}}{{n!}} $ and.
solution: In fact,
$$\begin{align*}{b_k} &= \sum\limits_{n =0}^\infty {\frac{{{n^k}}}{{n!}}} = \sum\limits_{n = 1}^\infty {\frac{{ {n^{k-1}}}} {{\left ({n-1} \right)!}}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left ({n + 1} \right)}^{k-1}}}}{{n!}} \\&= {b_{k-1}} + c_{k-1}^1{b_{k-2 }} + \cdots + c_{k-1}^{k-2}{b_1} + {b_0},\end{align*}$$
where $ b_0=e $.
The resulting number is called the bell number, recorded as $ B_n $, and
\[b\left (x \right) = \sum\limits_{n = 0}^\infty{\frac{{b\left (n \right)}}{{n!}} {X^n}} = {E^{{e^x}-1}}.\]

back to the original question, we have \[\sum\limits_{n = 0}^\infty {\frac{{p\left (n\right)}}{{n!}}} = E\sum\limits_{k = 0}^m {{a_k}{b_k}}. \]

3. Ask for $1-\frac{{{2^3}}}{{1!}} + \frac{{{3^3}}}{{2!}}-\frac{{{4^3}}}{{3!}} + \cdots $ .
solution: In fact,
$$\begin{align*}{b_k} &= \sum\limits_{n =0}^\infty {{{\left ({-1} \right)}^n}\frac{{{n^k}}}{{n!}} = \sum\ Limits_{n =1}^\infty {{\left ({-1} \right)}^n}\frac{{{n^{k-1}}}}{{\left ({n-1}\right)!}} = \sum\limits_{n = 0}^\in Fty {{{\left ({-1}\right)}^{n+1}}\frac{{{{\left ({n + 1} \right)}^{k-1}}}}{{n!}}} \\& =-{b_{k-1}}-c_{k-1}^1{ B_{k-2}}-\cdots-c_{k-1}^{k-2}{b_1}-{b_0},\end{align*}$$
where $ b_0=1/e $. So $ b_1=-1/e,b_2=0,b_3=1/e $.

so
$$\begin{align*}& 1-\frac{{{2^3}}}{{1!}} +\frac{{{3^3}}}{{2!}}-\frac{{{4^3}}}{{3!}} + \cdots = \sum\limits_{ N =0}^\infty {{{\left ({-1} \right)}^n}\frac{{{{\left ({n + 1}\right)}^3}}}{{n!}}} \\=& {b_3} + 3{b_2} + 3{b_1} + {b _0} =-\frac{1}{e}.\end{align*}$$

4. The sum of the following series: (1) $ \sum\limits_{n= 1}^\infty {\arctan \frac{1}{{2{n^2}}} $; (2) $ \sum\limits_{n = 1}^\infty{\arctan \frac{2}{{{n^2}}} $.
solution: In fact
\[\sum\limits_{n = 1}^\infty {\arctan\frac{1}{{2{n^2}}} = \sum\limits_{n = 1}^\infty {\left ({\arctan \frac{1}{{2n- 1}}-\arctan \frac{1}{{2n + 1}}} \right)} = \frac{\pi}{4}.\]
and
\[\sum\limits_{n = 1}^\infty {\arctan \frac{2}{{{n^2}}}}= \sum\limits_{n = 1}^\infty {\left ({\arctan \frac{1}{{n-1} }-\arctan\frac{1}{{n + 1}}} \right)} = \frac{\pi}{2} + \frac{\pi}{4} = \frac{{3\pi}}{4}.\]
5. Set $ A>1 $, ask $ \sum\limits_{n = 0}^\infty {\frac{{{2^n}}}{{{a^{{2^n}}} +1}} $ and.
solution: In fact
$$\begin{align*}\sum\limits_{n = 0}^\infty{\frac{{{2^n}}}{{{a^{{2^n}} + 1}}} &= \frac{1}{{a + 1} + \sum\limits _{n =1}^\infty {\frac{{{2^n}}}{{{a^{{2^n}} + 1}} = \frac{1}{{a + 1}}-\frac{1}{{a-1} + \frac{1}{{a + 1}} + \sum\limit S_{n = 1}^\infty {\frac{{{2^n}}}{{{a^{{2^n}}}+ 1}}} \\&= \frac{1}{{a + 1}}-\frac{2}{{{a^2}-1}} + \sum\limits_{n =1 }^\infty {\frac{{{2^n}}}{{{a^{{2^n}} + 1}}} = \frac{1}{{a + 1}}-\frac{{{2^2}}}{{{a^{{2^2}}-1}} + \sum\limits_{n = 2}^ \infty{\frac{{{2^n}}}{{{a^{{2^n}} + 1}}} \\&= \frac{1}{{a + 1}}-\mathop {\lim}\limits_{n \to \infty} \frac{{{2^{n + 1}}}}{{{a^{{2^{n + 1}}}-1}} =\frac{1}{{a + 1}}.\end{align*}$$

6. Request $1 +\frac{1}{3}-\frac{1}{5}-\frac{1}{7} + \frac{1}{9} + \frac{1}{{11}}-the and of \cdots$.
Solution:
$$\begin{align*}&\sum\limits_{n = 1}^\infty{\left ({\frac{1}{{8n-7}} + \frac{1}{{8n-5}}-\frac{1}{{8n-3}}- \frac{1}{{8n-1}}} \right)} = \sum\limits_{n = 1}^\infty {\int_0^1 {\left ({{x^{8n-8}} + {x^{8n-6}}-{x^{8n-4}}-{ X^{8n-2}}} \right)}} \\=&\int_0^1 {\sum\limits_{n = 1}^\infty {\left ({{x^{8n-8}} + {x^{8n-6}}-{x^{8n-4}}- {X^{8n-2}}} \right)} dx} = \int_0^1 {\frac{{1 + {x^2}-{x^4}-{x^6}}}{{1-{X^8}}}DX} \\= &\left. {\frac{{\arctan \left ({1 + \sqrt 2x} \right)-\arctan \left ({1-\sqrt 2 x} \right)}}{{\sqrt 2}} \right|_0^1= \frac{\ Pi}{{2\sqrt 2}}.\end{align*}$$

7. Ask for $1-\frac{1}{7} + \frac{1}{9}-\frac{1}{{15}} + \cdots $ for the and.
Solution:
$$\begin{align*}&\sum\limits_{n = 1}^\infty{\left ({\frac{1}{{8n-7}}-\frac{1}{{8n-1}}} \right)} = \sum\limit S_{n =1}^\infty {\int_0^1 {\left ({{x^{8n-8}}-{x^{8n-2}}} \right)}} \\=&\int_0^1 {\sum\limits_{n = 1}^\infty {\ Left ({{x^{8n-8}}-{x^{8n-2}}}\right)} dx} = \int_0^1 {\frac{{1-{x^6}}}{{1-{X^8}}}DX} \\= &\left. {\frac{{2\arctan x + \sqrt 2 \arctan \left ({1 + \sqrt 2 x} \right)-\arctan\left ({1-\sqrt 2 x} \right)}}{4}} \right|_ 0^1 = \frac{{\sqrt 2 + 1}}{8}\pi. \end{align*}$$

8. Ask for $1-\frac{1}{4} + \frac{1}{7}-\frac{1}{{10}} + \cdots $ for the and.
Solution:
$$\begin{align*}&\sum\limits_{n = 1}^\infty{\left ({\frac{1}{{6n-5}}-\frac{1}{{6n-2}}} \right)} = \sum\limit S_{n =1}^\infty {\int_0^1 {\left ({{x^{6n-6}}-{x^{6n-3}}} \right)}} \\=&\int_0^1 {\sum\limits_{n = 1}^\infty {\ Left ({{x^{6n-6}}-{x^{6n-3}}}\right)} dx} = \int_0^1 {\frac{{1-{x^3}}}{{1-{X^6}}}DX} = \int_0^1{\frac{1}{{1 + {x ^3}}}DX} \\=& \left. {\left ({-\frac{1}{6}\ln \left ({{x^2}-X + 1} \right) + \frac{1}{3}\ln \left ({x + 1} \right) +\frac{{\arctan \frac{{2x -1}}{{\sqrt 3}}}}{{\sqrt 3}} \right)} \right|_0^1= \frac{{\sqrt 3 \pi + 3\ln 2}}{9}.\end{align*}$$

9. Set $ {A_n} = 1 +\frac{1}{2} + \cdots + \frac{1}{n},n =, \cdots $, ask $ \sum\limits_{n = 1}^\infty {\frac{{{a_n}}}{ {N\left ({n +1} \right)}}} $ 's and.
Solution:
$$\begin{align*}&\sum\limits_{n = 1}^\infty{\frac{{{a_n}}}{{n\left ({n + 1} \right)}} = \sum\limits_{n = 1}^\ Infty{\frac{{1 + \frac{1}{2} + \cdots + \frac{1}{n}}}{{n\left ({n + 1} \right)}}}\\=&\sum\limits_{n = 1}^\infty {\left ({\frac{{1 + \frac{1}{2} + \cdots +\frac{1}{n}}}{n}-\frac{{1 + \frac{1}{2} + \cdots + \frac{1}{{n + 1}}}}{{n +1}}} \rig HT)} + \sum\limits_{n = 1}^\infty {\frac{1}{{{{\left ({n + 1}\right)}^2}}} \\= & 1-\mathop {\lim}\limits_{n \to \i NFTY} \frac{{1 +\frac{1}{2} + \cdots + \frac{1}{{n + 1}}}}{{n + 1} + \left ({\frac{{{\pi^2}}}{6}-1} \right) = \frac{{{ \pi ^2}}}{6}-\mathop {\lim}\limits_{n \to\infty} \frac{{\frac{1}{{n + 2}}}}{1} = \frac{{{\pi ^2}}}{6}.\end{align*}$$

10. Ask $ \sum\limits_{n = 0}^\infty {\left ({\frac{1}{{4n + 1}} +\frac{1}{{4n + 3}}-\frac{1}{{2n + 2}}} \right)} $ and.
Solution:
$$\begin{align*}&\sum\limits_{n = 0}^\infty{\left ({\frac{1}{{4n + 1}} + \frac{1}{{4n + 3}}-\frac{1}{{2n + 2}} \right)}= \sum\limits_{n = 0}^\infty {\int_0^1 {\left ({{x^{4n}} + {x^{4n + 2}}-{x^{2n + 1}} \right)}} \\= &\int_0^ 1 {\sum\limits_{n = 0}^\infty {\left ({{x^{4n}} + {x^{4n + 2}}-{x^{2n + 1}}} \right)} dx} = \int_0^1 {\left ({\frac{{1 + { X^2}}}{{1-{X^4}}}-\frac{x}{{1-{x^2}}}} \right) dx} \\=&\int_0^1 {\frac{1}{{1 + x}}dx} = \ln 2.\end{align*}$$

11. Ask for $1-\frac{1}{4} + \frac{1}{6}-\frac{1}{9} + \frac{1}{{11}}-\frac{1}{{14}} + \cdots $ .
Solution:
$$\begin{align*}&\sum\limits_{n = 1}^\infty{\left ({\frac{1}{{5n-4}}-\frac{1}{{5n-1}}} \right)} = \sum\limit S_{n =1}^\infty {\int_0^1 {\left ({{x^{5n-5}}-{x^{5n-2}}} \right) dx}} \\=&\int_0^1 {\sum\limits_{n = 1}^\infty {\left ({{x^{5n-5}}-{x^{5n-2}}}\right)} dx} = \int_0^1 {\frac{{1-{x^3}}}{{1-{X^5}}}DX} \\= &\left. {\left ({\frac{{\left ({5-\sqrt 5} \right)/10}}{{{x^2} + \frac{{\sqrt 5 +1}}{2}x + 1}} + \frac{{\left ({5 + \sqrt 5} \ right)/10}}{{{x^2} + \frac{{-\sqrt 5 + 1}}{2}x + 1}}} \right)} \right|_0^1 = \frac{{\sqrt {+ 10\sqrt 5}}}{{25}}\pi. \e nd{align*}$$

12. Ask for $ \frac{{{x^3}}}{{3!}} + \frac{{{x^9}}}{{9!}} +\frac{{{x^{15}}}}{{15!}} + \cdots $ and function.
solution: In fact, the equation $ \omega^3=1$ has three roots $1,{-\frac{1}{2} +\frac{{\sqrt 3 i}}{2}},{-\frac{1}{2}-\frac{{\sqrt 3 i}}{2}} $. Using $ \sinh $ to get the required function
$$\begin{align*}&\frac{{\sinh x + \sinh \left ({-\frac{1}{2} + \frac{{\sqrt 3 i}}{2}} \right) x + \sinh \left ({- \FRAC{1}{2}-\frac{{\sqrt 3 i}}{2}} \right) x}}{3}\\= &-\frac{2}{3}\sinh\frac{x}{2}\cos \frac{{\sqrt 3 x}}{2} + \fra C{{\sinh x}}{3} =\frac{{{x^3}}}{{3!}} + \frac{{{x^9}}}{{9!}} + \frac{{{x^{15}}}}{{15!}} + \cdots.\end{align*}$$

We also have

$$\begin{align*}&{\frac{{\sin x +\sin \left ({-\frac{1}{2} + \frac{{\sqrt 3 i}}{2}} \right) x + \sin \left ({-\f RAC{1}{2}-\frac{{\sqrt 3 i}}{2}} \right) x}}{{-3}}}\\= &\frac{2}{3}\sin\frac{x}{2}\cosh \frac{{\sqrt 3 x}}{2}-\fra C{{\sin x}}{3} = \frac{{{x^3}}}{{3!}} -\frac{{{x^9}}}{{9!}} + \frac{{{x^{15}}}}{{15!}}-\frac{{{x^{21}}}}{{21!}} +\cdots. \end{align*}$$

13. Ask $ \sum\limits_{n = 1}^\infty {\frac{{{{\left[{\left {n-1}\right)!} \right]}^2}}}{{\left ({2n} \right)!}} {{\left ({2x} \right)}^{2n}}} $ and the function.
solution: On $ |x|<1 $ on the derivative of $ s (x) $, know $ S ' \left (x \right) = 2\sum\limits_{n = 1}^\infty{\frac{{{{\left[{\left ({n-1} \right)!} \right]}^2}}}{{\left ({2n-1}\right)!}} {{\left ({2x} \right)}^{2n-1}}} $, and $ S ' \left (x \right) = 4\sum\limits_{n = 1}^\infty{\frac{{{{\left[{\left ({n-1} \right)!} \right]}^2}}}{{\left ({ 2n-2}\right)!}} {{\left ({2x} \right)}^{2n-2}}} $. This may be $ (1-x^2) S ' (x)-xs ' (x) =4 $. At both ends multiplied by $ {(1-x^2)}^{-1/2} $, we have
\[{\left ({\sqrt {1-{x^2}} S ' \left (x \right)}\right) ^\prime} = \frac{4}{{\sqrt {1-{x^2}}}},\]
\[s\left (x \right) = \frac{{4\arcsin x}}{{\sqrt{1-{x^2}}} + \frac{1}{{\sqrt {1-{x^2}}}},\quad \left| x \right| < 1.\]

14. Ask $ \sum\limits_{n = 1}^\infty {\frac{{{x^{n + 1}}}}{{\left ({1-{x^n}} \right) \left ({1-{x^{n + 1}}}}}} $ and the function.
the solution: notice
$$\begin{align*}&\left ({1-\frac{1}{x}}\right) \sum\limits_{n = 1}^\infty {\frac{{{x^{n + 1}}}}{{\left ({1-{x^ N}}\right) \left ({1-{x^{n + 1}} \right)}} \\=& \sum\limits_{n = 1}^\infty{\frac{{{x^{n + 1}}}}{{\left ({1-{x^n}} \right) \left ({1-{x^{n + 1}}}\right)}}-\sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{\left ({1-{x^n}}\right) \left ({ 1-{x^{n + 1}} \right)}}} \\= &\sum\limits_{n = 1}^\infty{\frac{{{x^{n + 1}}-{X^n}}}{{\left ({1-{x^n}} \right) \l EFT ({1-{x^{n + 1}}}\right)}}} = \sum\limits_{n = 1}^\infty {\left ({\frac{1}{{1-{x^{n + 1}}}}-\frac{1}{{1-{x^n}}} \right)} \\=& \mathop {\lim}\limits_{n \to \infty}\frac{1}{{1-{x^{n + 1}}}-\frac{1}{{1-x}} = \begin{cases}\f Rac{1}{{x-1}},&\left| X \right| > 1\\\frac{x}{{x-1}},&\left| X \right| <1\end{cases}. \end{align*}$$
so
\[\sum\limits_{n = 1}^\infty {\frac{{{x^{n +1}}}}{{\left ({1-{x^n}} \right) \left ({1-{x^{n + 1}}} \right)}} =\beg In{cases}\frac{x}{{{{\left ({x-1} \right)}^2}}, &\left| X \right|> 1\\\frac{{{x^2}}}{{{{\left ({x-1} \right)}^2}}, &\left| x \right|< 1\end{cases}. \]

15. Set $ \sum\limits_{n = 1}^\infty {\frac{1}{{{a_n}}} $ for divergent positive progression, $ x>0 $, $ \sum\limits_{n = 1}^\infty {\frac{{{ a_1}{a_2} \cdots{a_n}}}{{\left ({{a_2} + x} \right) \cdots \left ({{a_{n + 1}} + x} \right)}}} $ and function.
solution: First,
$$\begin{align*}&\sum\limits_{n = 1}^\infty{\frac{{{a_1}{a_2} \cdots {A_n}}}{{\left ({{a_2} + x} \right) \cdots \left ({{a_{n + 1}} + x} \right)}} \\=& \frac{{{a_1}}}{{{a_2} + x}} +\frac{1}{x}\sum\limits_{n = 2}^\infty {\left[{\f rac{{{a_1}{a_2} \cdots{a_n}}}{{\left ({{a_2} + x} \right) \cdots \left ({{a_n} + x} \right)}}-\frac{{{a_1}{a_2} \cdots {A _{n + 1}}}}{{\left ({{a_2} + x} \right) \cdots\left ({{a_{n + 1}} + x} \right)}}} \right]} \\=& \frac{{{a_1}}}{{{a_2} +X}} + \frac{1}{x}\left[{\frac{{{a_1}{a_2}}}{{{a_2} + x}}-\mathop {\lim}\limits_{n \to \infty} \frac{{{a_1}{a_2} \cdot s {a_{n + 1}}}}{{\left ({{a_2}+ x} \right) \cdots \left ({{a_{n + 1}} + x} \right) }}} \right].\end{align*}$$
when $ N $ is large enough, \[1 + \frac{x}{{{a_{n + 1}}} \sim {e^{x/{a_{n +1}}}}.\]
therefore $ {\left ({1+ \frac{x}{{{a_2}}} \right) \cdots \left ({1 + \frac{x}{{{a_{n + 1}}}}}\right)} $ with \exp \left\{{x\s Um\limits_{n = 1}^\infty {\frac{1}{{{a_n}}}} \right\} $ has the same convergence, all divergent, so
\[\mathop {\lim}\limits_{n \to \infty}\frac{{{a_1}{a_2} \cdots {a_{n + 1}}}}{{\left ({{a_2} + x} \right) \cdots\left ({{a_{n + 1}} + x} \right}} = \mathop {\lim}\limits_{n \to \infty}\frac{{{a_1}}}{{\left ({1 + \frac{x}{{{a_2}}} \right) \cdots \left ({1 + \frac{x} {{{a_{n+ 1}}}} \right)}} = 0.\]
thereby
\[\sum\limits_{n = 1}^\infty {\frac{{{a_1}{a_2}\cdots {a_n}}}{{\left ({{a_2} + x} \right) \cdots \left ({{a_{n + 1}} + x}\right)}}} = \frac{{{a_1}}}{{{a_2} + x}} + \frac{{{a_1}{a_2}}}{{x\left ({{a_2}+ x} \right)}} = \frac{{{a_1}}}{x}.\]

16. Set $ X>1 $, request $ \frac{x}{{x+ 1} + \frac{{{x^2}}}{{\left ({x + 1} \right) \left ({{x^2} + 1} \right)}} +\frac{{{x^ 4}}}{{\left ({x + 1} \right) \left ({{x^2} + 1} \right) \left ({{x^4} +1} \right)} + \cdots $ and functions.
solution: $$\begin{align*}i&= \left ({1-\frac{1}{{x + 1}}} \right) + \frac {{{x^2}}} {{\left ({x + 1}\right) \left ({{x^2} + 1} \right)} + \frac{{{x^4}}}{{\left ({x + 1}\right) \left ({{x^2} + 1} \right) \left ({{x^4} + 1} \right)}} + \cdots \\&=1 + \left ({-\frac{1}{{x + 1}} + \frac{{{x^2}}}{{\left ({x + 1} \right) \left ({{x^2} + 1} \right)}}} \ri ght) + \frac{{{x^4}}}{{\left ({x + 1} \right) \left ({{x^2} + 1} \right) \left ({{x^4} + 1} \right)}} + \cdots \\&= 1-\f Rac{1}{{\left ({x + 1} \right) \left ({{x^2} + 1} \right)}} + \frac{{{x^4}}}{{\left ({x + 1} \right) \left ({{x^2} + 1} \righ T) \left ({{x^4} + 1} \right)}} + \cdots\\&= 1-\frac{1}{{\left ({x + 1} \right) \left ({{x^2} + 1} \right) \left ({{x^4} + 1} \right}} + \cdots \\&= \cdots = 1-\mathop {\lim}\limits_{n\to \infty} \frac{1}{{\left ({x + 1} \right) \left ({{x^2} + 1} \right) \cdots\left ({{x^{{2^{n-1}}} + 1} \right)}} = 1.\end{align*}$$ 

Originating from: http://www.math.org.cn/forum.php?mod=viewthread&tid=35174 [not verified its correctness, for reference only]

Sheihuimin, Yun, Yi-Huai, Chanding-bian compilation of mathematical Analysis exercises Lecture notes section 16.2 exercises reference answers [from Tao younger brother]