Convergence Tests

$\lim_{n\to\infty} \dfrac{4 - 1/n}{7 + 2/n^{2}} = \dfrac{4}{7}$.

$\dfrac{4}{7} \neq 0$.

Diverges by the divergence test.

$\dfrac{n}{n+1} \to 1$, so the terms swing between values near $1$ and $-1$: $\lim_{n\to\infty} a_n$ does not exist.

No limit, so certainly not $0$.

Diverges by the divergence test.

With $t = \tfrac{1}{n} \to 0$: $n\sin\dfrac{1}{n} = \dfrac{\sin t}{t} \to 1$.

$1 \neq 0$.

Diverges by the divergence test.

$\dfrac{1}{\sqrt{n}} \to 0$.

The limit is $0$, so the test says nothing.

Inconclusive. Another test decides it: a $p$-series with $p = \tfrac{1}{2} \leq 1$, so it diverges.

$f(x) = x e^{-x^{2}}$.

$f > 0$ and continuous for $x \geq 1$; $f'(x) = e^{-x^{2}}(1 - 2x^{2}) < 0$ for $x \geq 1$.

$\int_1^{\infty} x e^{-x^{2}}\,dx = \lim_{t\to\infty}\left[-\tfrac{1}{2}e^{-x^{2}}\right]_1^{t} = \dfrac{1}{2e}$.

The integral converges, so the series converges.

$f(x) = \dfrac{1}{x\sqrt{\log x}}$.

For $x \geq 2$: positive and continuous, and decreasing because $x$ and $\sqrt{\log x}$ both increase.

With $u = \log x$, $du = \dfrac{dx}{x}$: $\int_2^{t} \dfrac{dx}{x\sqrt{\log x}} = 2\sqrt{\log t} - 2\sqrt{\log 2} \to \infty$.

The integral diverges, so the series diverges.

$f(x) = \dfrac{\log x}{x^{2}}$.

For $x \geq 2$, $f$ is positive and continuous. $f'(x) = \dfrac{1 - 2\log x}{x^{3}} < 0$ once $x > \sqrt{e} \approx 1.65$, so $f$ decreases there.

By parts: $\int_2^{t} \dfrac{\log x}{x^{2}}\,dx = \left[-\dfrac{\log x + 1}{x}\right]_2^{t} \to \dfrac{\log 2 + 1}{2}$.

The integral converges, so the series converges.

$a_n \approx \dfrac{\sqrt{n}}{n^{2}} = \dfrac{1}{n^{3/2}}$, so $b_n = \dfrac{1}{n^{3/2}}$.

$p = \tfrac{3}{2} > 1$: $\sum b_n$ converges.

$\dfrac{a_n}{b_n} = \dfrac{n^{3/2}(\sqrt{n} + 2)}{n^{2} + 5} = \dfrac{n^{2} + 2n^{3/2}}{n^{2} + 5} \to 1$.

$L = 1$, so the series converges by the limit comparison test.

$3^{n}$ beats $n^{2}$ and $4^{n}$ beats $1$: $b_n = \left(\dfrac{3}{4}\right)^{n}$.

Geometric with $r = \tfrac{3}{4} < 1$: $\sum b_n$ converges.

$\dfrac{a_n}{b_n} = \dfrac{4^{n}(3^{n} + n^{2})}{3^{n}(4^{n} - 1)} = \dfrac{1 + n^{2}/3^{n}}{1 - 1/4^{n}} \to 1$.

$L = 1$, so the series converges by the limit comparison test.

The top stays between $1$ and $3$, so $a_n$ behaves like $\dfrac{1}{n^{3}}$; take $b_n = \dfrac{3}{n^{3}}$.

$p = 3 > 1$: $\sum \dfrac{3}{n^{3}}$ converges.

$\cos n \leq 1$, so $0 < \dfrac{2 + \cos n}{n^{3}} \leq \dfrac{3}{n^{3}}$ for all $n \geq 1$.

Smaller than a convergent series: converges by the comparison test.

$\log n$ grows more slowly than $n$, so compare with $b_n = \dfrac{1}{n}$.

Harmonic series: $\sum \dfrac{1}{n}$ diverges.

$\log n < n$ for $n \geq 2$, so $\dfrac{1}{\log n} > \dfrac{1}{n} > 0$.

Bigger than a divergent series: diverges by the comparison test.

$a_{n+1} = \dfrac{(n+1)^{2}}{3^{n+1}}$.

$\dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)^{2}}{3^{n+1}} \cdot \dfrac{3^{n}}{n^{2}} = \dfrac{1}{3}\left(\dfrac{n+1}{n}\right)^{2}$.

$L = \dfrac{1}{3}$.

$L < 1$: converges by the ratio test.

$a_{n+1} = \dfrac{(n+1)!}{(n+1)^{2}\,2^{n+1}}$.

$\dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)!}{n!} \cdot \dfrac{n^{2}}{(n+1)^{2}} \cdot \dfrac{2^{n}}{2^{n+1}} = \dfrac{n^{2}}{2(n+1)}$.

$L = \infty$.

$L > 1$: diverges by the ratio test.

$a_{n+1} = \dfrac{(2n+2)!}{(n+1)!\,(n+1)^{n+1}}$.

$\dfrac{a_{n+1}}{a_n} = \dfrac{(2n+2)(2n+1)}{n+1} \cdot \dfrac{n^{n}}{(n+1)^{n+1}} = \dfrac{2(2n+1)}{n+1}\left(\dfrac{n}{n+1}\right)^{n}$.

$\left(\dfrac{n}{n+1}\right)^{n} = \dfrac{1}{(1 + 1/n)^{n}} \to \dfrac{1}{e}$, so $L = \dfrac{4}{e} \approx 1.47$.

$L > 1$: diverges by the ratio test.

$|a_n| = \dfrac{1}{\sqrt{n} + 3}$; limit comparison with $\dfrac{1}{\sqrt{n}}$ gives $L = 1$, and $p = \tfrac{1}{2} \leq 1$: $\sum |a_n|$ diverges.

Not absolutely convergent.

$b_n = \dfrac{1}{\sqrt{n} + 3}$ decreases (the bottom grows) and tends to $0$.

Converges by the alternating series test, so it is conditionally convergent.

$|a_n| = \dfrac{n^{2}}{n^{4} + 1} < \dfrac{n^{2}}{n^{4}} = \dfrac{1}{n^{2}}$, and $\sum \dfrac{1}{n^{2}}$ converges: $\sum |a_n|$ converges.

Absolutely convergent. Done.

Skipped: absolute convergence already settles it.

Absolutely convergent, so it converges.

$|a_n| = \dfrac{2n + 1}{3n - 1} \to \dfrac{2}{3} \neq 0$: $\sum |a_n|$ diverges.

Not absolutely convergent.

$b_n \to \dfrac{2}{3}$, not $0$: the alternating series test fails.

$a_n$ has no limit (it swings near $\pm\tfrac{2}{3}$), so the series diverges by the divergence test.

$|a_n| = \dfrac{|\cos 2n|}{n^{3/2}} \leq \dfrac{1}{n^{3/2}}$, and $p = \tfrac{3}{2} > 1$: $\sum |a_n|$ converges by comparison.

Absolutely convergent. Done.

Skipped: the signs of $\cos(2n)$ ($-, -, +, -, \dots$) do not alternate anyway.

Absolutely convergent, so it converges.

$b_n = \dfrac{1}{n^{4}}$ is positive, decreasing and tends to $0$.

$|S - S_N| \leq \dfrac{1}{(N+1)^{4}}$.

$\dfrac{1}{(N+1)^{4}} \leq 10^{-4}$ means $(N+1)^{4} \geq 10^{4}$, so $N + 1 \geq 10$ and $N = 9$.

$b_n = \dfrac{1}{n!}$ is positive, decreasing and tends to $0$.

$|S - S_N| \leq \dfrac{1}{(N+1)!}$.

$6! = 720 < 1000$ but $7! = 5040 \geq 1000$, so $N + 1 = 7$ and $N = 6$.

$b_n = \dfrac{1}{n^{2}\,2^{n}}$ is positive, decreasing (both factors grow) and tends to $0$.

$|S - S_N| \leq \dfrac{1}{(N+1)^{2}\,2^{N+1}}$.

Need $(N+1)^{2}\,2^{N+1} \geq 100$: $N + 1 = 3$ gives $72$, $N + 1 = 4$ gives $256$. So $N = 3$.