## 2019年8月5日月曜日

### 数学 - Python - 解析学 - 級数 - 比による判定法 - 対数関数、逆数、発散

1. $\begin{array}{l}\frac{{a}_{n+1}}{{a}_{n}}\\ =\frac{{\left(\mathrm{log}n\right)}^{10}}{{\left(\mathrm{log}\left(n+1\right)\right)}^{10}}\\ ={\left(\frac{\mathrm{log}n}{\mathrm{log}\left(n+1\right)}\right)}^{10}\\ \frac{\frac{d}{dn}\mathrm{log}n}{\frac{d}{dn}\mathrm{log}\left(n+1\right)}\\ =\frac{n+1}{n}\\ =1+\frac{1}{n}\end{array}$

よって、

$\underset{n\to \infty }{\mathrm{lim}}\frac{\mathrm{log}n}{\mathrm{log}\left(n+1\right)}=1$

ゆえに、

$\begin{array}{l}\frac{{a}_{n+1}}{{a}_{n}}

を満たす c は存在しないので、問題の無限級数は発散する。

コード

Python 3

#!/usr/bin/env python3
from sympy import pprint, symbols, plot, summation, oo, log, Limit
import matplotlib.pyplot as plt

print('13.')

n = symbols('n', integer=True)
s = summation(1 / log(n) ** 10, (n, 2, oo))
l = Limit(log(n) ** 10 / log(n + 1) ** 10, n, oo)

for o in [s, l, l.doit()]:
pprint(o)
print()

def f(n):
return sum([1 / log(k) ** 10 for k in range(2, n + 1)])

ns = range(2, 20)
plt.plot(ns, [f(n) for n in ns],
ns, [log(n) ** 10 / log(n + 1) ** 10 for n in ns],
ns, [1 for _ in ns])

plt.legend(['Σ 1 / log n', 'an + 1 / an', 1])
plt.savefig('sample13.png')


C:\Users\...>py sample13.py
13.
∞
____
╲
╲      1
╲  ────────
╱     10
╱   log  (n)
╱
‾‾‾‾
n = 2

⎛     10     ⎞
⎜  log  (n)  ⎟
lim ⎜────────────⎟
n─→∞⎜   10       ⎟
⎝log  (n + 1)⎠

1

c:\Users\...>