1 − 1 2 + 1 3 − 1 4 + . . . + ( − 1 ) n − 1 n + . . . ( ∗ ) a n = ( − 1 ) n − 1 n = ( − 1 ) n − 1 n 1 2 . 1-\frac{1}{\sqrt2}+\frac{1}{\sqrt3}-\frac{1}{\sqrt4}+...+\frac{(-1)^{n-1}}{\sqrt n}+...(*)\\
a_n=\frac{(-1)^{n-1}}{\sqrt n}=\frac{(-1)^{n-1}}{ n^\frac{1}{2}}. 1 − 2 1 + 3 1 − 4 1 + ... + n ( − 1 ) n − 1 + ... ( ∗ ) a n = n ( − 1 ) n − 1 = n 2 1 ( − 1 ) n − 1 .
Consider a series of modules
1 + 1 2 + 1 3 + 1 4 + . . . + 1 n + . . . ( ∗ ∗ ) 1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}+\frac{1}{\sqrt4}+...+\frac{1}{\sqrt n}+...(**)\\ 1 + 2 1 + 3 1 + 4 1 + ... + n 1 + ... ( ∗ ∗ )
Series
∑ n = 1 ∞ 1 n α \sum\limits_{n=1}^{\infty} \frac{1}{n^\alpha} n = 1 ∑ ∞ n α 1
convergent if α > 1 \alpha>1 α > 1 , divergent if α ≤ 1 \alpha \leq1 α ≤ 1 .
In our case α = 1 2 < 1 \alpha=\frac{1}{2}<1 α = 2 1 < 1 then series (**) is divergent.
We examine the series (*) on the basis of Leibniz
1 ) ∣ a 1 ∣ > ∣ a 2 ∣ > . . . > ∣ a n ∣ > . . . 1) |a_1|>|a_2|>...>|a_n|>... 1 ) ∣ a 1 ∣ > ∣ a 2 ∣ > ... > ∣ a n ∣ > ...
in our case
1 > 1 2 > 1 3 > . . . > 1 n > . . . 2 ) lim n ↦ ∞ ∣ a n ∣ = lim n ↦ ∞ 1 n = 0. 1>\frac{1}{\sqrt 2}>\frac{1}{\sqrt 3}>...>\frac{1}{\sqrt n}>...\\
2) \lim\limits_{n\mapsto \infty}|a_n|=\lim\limits_{n\mapsto \infty} \frac{1}{\sqrt n}=0. 1 > 2 1 > 3 1 > ... > n 1 > ... 2 ) n ↦ ∞ lim ∣ a n ∣ = n ↦ ∞ lim n 1 = 0.
According to Leibniz, the series (*) is convergent.
Since the series (**) is divergent and the series (*) is convergent, the series (*) is conventionally convergent.
Comments