5. -/1.42 points details my notes test the series for convergence or divergence. ∑n = 1∞(−1)nn9−1n10+1…

Explanation:

Step1: Recall the Alternating - Series Test

The given series is $\sum_{n = 1}^{\infty}(-1)^{n}\frac{n^{9}-1}{n^{10}+1}$, which is an alternating series of the form $\sum_{n = 1}^{\infty}(-1)^{n}a_{n}$, where $a_{n}=\frac{n^{9}-1}{n^{10}+1}$.

Step2: Check the two conditions of the Alternating - Series Test

Condition 1: $\lim_{n\rightarrow\infty}a_{n}$

We find $\lim_{n\rightarrow\infty}a_{n}=\lim_{n\rightarrow\infty}\frac{n^{9}-1}{n^{10}+1}$. Divide both the numerator and denominator by $n^{10}$: [ \begin{align*} \lim_{n\rightarrow\infty}\frac{n^{9}-1}{n^{10}+1}&=\lim_{n\rightarrow\infty}\frac{\frac{n^{9}}{n^{10}}-\frac{1}{n^{10}}}{\frac{n^{10}}{n^{10}}+\frac{1}{n^{10}}}\ &=\lim_{n\rightarrow\infty}\frac{\frac{1}{n}-\frac{1}{n^{10}}}{1 + \frac{1}{n^{10}}}\ &= 0 \end{align*} ]

Condition 2: $a_{n+1}\leq a_{n}$ for all $n$ large enough

Let $f(x)=\frac{x^{9}-1}{x^{10}+1}$, then find its derivative using the quotient - rule. The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = x^{9}-1$, $u^\prime=9x^{8}$, $v=x^{10}+1$, $v^\prime = 10x^{9}$. [ \begin{align*} f^\prime(x)&=\frac{9x^{8}(x^{10}+1)-(x^{9}-1)\times10x^{9}}{(x^{10}+1)^{2}}\ &=\frac{9x^{18}+9x^{8}-10x^{18}+10x^{9}}{(x^{10}+1)^{2}}\ &=\frac{-x^{18}+10x^{9}+9x^{8}}{(x^{10}+1)^{2}}\ &=\frac{-x^{8}(x^{10}-10x - 9)}{(x^{10}+1)^{2}} \end{align*} ] For $x$ large enough (say $x\geq1$), the function $y = x^{10}-10x - 9$ is positive. So, $f^\prime(x)<0$ for $x$ large enough. This means $a_{n + 1}\leq a_{n}$ for $n$ large enough. Since both conditions of the Alternating - Series Test are satisfied, the series $\sum_{n = 1}^{\infty}(-1)^{n}\frac{n^{9}-1}{n^{10}+1}$ is convergent.

Answer:

convergent