name: (1) (34 points) determine if the series is converging or diverging. (a) (8 points) $sum_{n =…

name: (1) (34 points) determine if the series is converging or diverging. (a) (8 points) $sum_{n = 1}^{infty}\frac{(ln n)^{3}+1}{n^{2}}$. (b) (6 points) $sum_{n = 1}^{infty}\frac{|cos(n + 1)|+3}{n + 2}$. (c) (12 points) $sum_{n = 1}^{infty}\frac{2n+cos(n^{3}-1)}{e^{n^{2}}}$. (d) (8 points) $sum_{n = 1}^{infty}\frac{n^{1.9}}{(ln n)^{3}}$.

name: (1) (34 points) determine if the series is converging or diverging. (a) (8 points) $sum_{n = 1}^{infty}\frac{(ln n)^{3}+1}{n^{2}}$. (b) (6 points) $sum_{n = 1}^{infty}\frac{|cos(n + 1)|+3}{n + 2}$. (c) (12 points) $sum_{n = 1}^{infty}\frac{2n+cos(n^{3}-1)}{e^{n^{2}}}$. (d) (8 points) $sum_{n = 1}^{infty}\frac{n^{1.9}}{(ln n)^{3}}$.

Answer

Explanation:

Step1: Recall the limit - comparison test

For a series (\sum_{n = 1}^{\infty}a_n) and a known - behavior series (\sum_{n = 1}^{\infty}b_n), we consider (\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=L), where (0\lt L\lt\infty). If (\sum_{n = 1}^{\infty}b_n) converges, then (\sum_{n = 1}^{\infty}a_n) converges, and if (\sum_{n = 1}^{\infty}b_n) diverges, then (\sum_{n = 1}^{\infty}a_n) diverges.

Step2: Analyze part (a)

Let (a_n=\frac{(\ln n)^3 + 1}{n^2}). For large (n), ((\ln n)^3\ll n^2). We use the fact that (\frac{(\ln n)^3+1}{n^2}\sim\frac{(\ln n)^3}{n^2}). We know that for any positive real numbers (p) and (q), (\lim_{n\rightarrow\infty}\frac{(\ln n)^p}{n^q}=0) when (q > 0). Also, we can use the limit - comparison test with (b_n=\frac{1}{n^{1.5}}) (a (p) - series with (p = 1.5>1)). (\lim_{n\rightarrow\infty}\frac{\frac{(\ln n)^3 + 1}{n^2}}{\frac{1}{n^{1.5}}}=\lim_{n\rightarrow\infty}\frac{(\ln n)^3 + 1}{n^{0.5}} = 0). Since (\sum_{n = 1}^{\infty}\frac{1}{n^{1.5}}) converges (by the (p) - series test, (\sum_{n = 1}^{\infty}\frac{1}{n^p}) converges for (p>1)), the series (\sum_{n = 1}^{\infty}\frac{(\ln n)^3 + 1}{n^2}) converges.

Step3: Analyze part (b)

We know that (|\cos(n + 1)|\leq1). So, (a_n=\frac{|\cos(n + 1)|+3}{n + 2}\leq\frac{1 + 3}{n+2}=\frac{4}{n + 2}). We use the limit - comparison test with the harmonic - like series (b_n=\frac{1}{n}). (\lim_{n\rightarrow\infty}\frac{\frac{|\cos(n + 1)|+3}{n + 2}}{\frac{1}{n}}=\lim_{n\rightarrow\infty}\frac{n(|\cos(n + 1)|+3)}{n + 2}). Since (|\cos(n + 1)|\in[0,1]), (\lim_{n\rightarrow\infty}\frac{n(|\cos(n + 1)|+3)}{n + 2}=\lim_{n\rightarrow\infty}\frac{|\cos(n + 1)|n+3n}{n + 2}). Dividing numerator and denominator by (n), we get (\lim_{n\rightarrow\infty}\frac{|\cos(n + 1)| + 3}{1+\frac{2}{n}}=3). Since (\sum_{n = 1}^{\infty}\frac{1}{n}) diverges (harmonic series), the series (\sum_{n = 1}^{\infty}\frac{|\cos(n + 1)|+3}{n + 2}) diverges.

Step4: Analyze part (c)

Let (a_n=\frac{2n+\cos(n^3)-1}{e^{n^2}}). We know that (|\cos(n^3)|\leq1), so (2n+\cos(n^3)-1\leq2n + 1 - 1=2n). Also, (e^{n^2}) grows much faster than any polynomial function of (n). We use the ratio test. Let (a_n=\frac{2n+\cos(n^3)-1}{e^{n^2}}), then (\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_n}\right|=\lim_{n\rightarrow\infty}\left|\frac{2(n + 1)+\cos((n + 1)^3)-1}{e^{(n + 1)^2}}\cdot\frac{e^{n^2}}{2n+\cos(n^3)-1}\right|). Since (e^{(n + 1)^2}=e^{n^2+2n + 1}=e^{n^2}\cdot e^{2n + 1}), (\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_n}\right|=\lim_{n\rightarrow\infty}\left|\frac{2(n + 1)+\cos((n + 1)^3)-1}{2n+\cos(n^3)-1}\cdot\frac{1}{e^{2n + 1}}\right| = 0). Since (\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|=0<1), the series (\sum_{n = 1}^{\infty}\frac{2n+\cos(n^3)-1}{e^{n^2}}) converges.

Step5: Analyze part (d)

Let (a_n=\frac{n^{1.9}}{(\ln n)^3}). We use the limit - comparison test with (b_n=\frac{1}{n^{0.1}}). (\lim_{n\rightarrow\infty}\frac{\frac{n^{1.9}}{(\ln n)^3}}{\frac{1}{n^{0.1}}}=\lim_{n\rightarrow\infty}\frac{n^{2}}{(\ln n)^3}=\infty). Since (\sum_{n = 1}^{\infty}\frac{1}{n^{0.1}}) diverges (a (p) - series with (p = 0.1<1)), the series (\sum_{n = 1}^{\infty}\frac{n^{1.9}}{(\ln n)^3}) diverges.

Answer:

(a) Converges (b) Diverges (c) Converges (d) Diverges