발산 판정법과 적분 판정법

In the following exercises, find a value of $N$ such that ${R}_{N}$ is smaller than the desired error. Compute the corresponding sum $\sum_{n=1}^{N}{a}_{n}$ and compare it to the given estimate of the infinite series.

${a}_{n}=\frac{1}{{n}^{11}},$ error $<{10}^{-4},$ $\sum_{n=1}^{\infty }\frac{1}{{n}^{11}}=1.000494\ldots$

In the following exercises, find a value of $N$ such that ${R}_{N}$ is smaller than the desired error. Compute the corresponding sum $\sum_{n=1}^{N}{a}_{n}$ and compare it to the given estimate of the infinite series.

${a}_{n}=\frac{1}{{e}^{n}},$ error $<{10}^{-5},$ $\sum_{n=1}^{\infty }\frac{1}{{e}^{n}}=\frac{1}{e-1}=0.581976\ldots$

In the following exercises, find a value of $N$ such that ${R}_{N}$ is smaller than the desired error. Compute the corresponding sum $\sum_{n=1}^{N}{a}_{n}$ and compare it to the given estimate of the infinite series.

${a}_{n}=\frac{n}{{e}^{{n}^{2}}},$ error $<{10}^{-5},$ $\sum_{n=1}^{\infty }n/{e}^{{n}^{2}}=0.40488139857\ldots$

In the following exercises, find a value of $N$ such that ${R}_{N}$ is smaller than the desired error. Compute the corresponding sum $\sum_{n=1}^{N}{a}_{n}$ and compare it to the given estimate of the infinite series.

${a}_{n}=1/{n}^{4},$ error $<{10}^{-4},$ $\sum_{n=1}^{\infty }1/{n}^{4}={\pi }^{4}/90=1.08232...$

In the following exercises, find a value of $N$ such that ${R}_{N}$ is smaller than the desired error. Compute the corresponding sum $\sum_{n=1}^{N}{a}_{n}$ and compare it to the given estimate of the infinite series.

${a}_{n}=1/{n}^{6},$ error $<{10}^{-6},$ $\sum_{n=1}^{\infty }1/{n}^{6}={\pi }^{6}/945=1.01734306...,$

Find the limit as $n\to \infty$ of $\frac{1}{n}+\frac{1}{n+1}+\cdots +\frac{1}{2n}.$ (Hint: Compare to ${\int }_{n}^{2n}\frac{1}{t}dt.)$

In the following exercises, find a value of $N$ such that ${R}_{N}$ is smaller than the desired error. Compute the corresponding sum $\sum_{n=1}^{N}{a}_{n}$ and compare it to the given estimate of the infinite series.

Find the limit as $n\to \infty$ of $\frac{1}{n}+\frac{1}{n+1}+\cdots +\frac{1}{3n}$

[T] Complete sampling with replacement, sometimes called the coupon collector’s problem, is phrased as follows: Suppose you have $N$ unique items in a bin. At each step, an item is chosen at random, identified, and put back in the bin. The problem asks what is the expected number of steps $E(N)$ that it takes to draw each unique item at least once. It turns out that $E(N)=N$. ${H}_{N}=N\left(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{N}\right)$. Find $E(N)$ for $N=10,20,\text{and}\,50$.

[T] The simplest way to shuffle cards is to take the top card and insert it at a random place in the deck, called top random insertion, and then repeat. We will consider a deck to be randomly shuffled once enough top random insertions have been made that the card originally at the bottom has reached the top and then been randomly inserted. If the deck has $n$ cards, then the probability that the insertion will be below the card initially at the bottom (call this card $B)$ is $1/n.$ Thus the expected number of top random insertions before $B$ is no longer at the bottom is n. Once one card is below $B,$ there are two places below $B$ and the probability that a randomly inserted card will fall below $B$ is $2/n.$ The expected number of top random insertions before this happens is $n/2.$ The two cards below $B$ are now in random order. Continuing this way, find a formula for the expected number of top random insertions needed to consider the deck to be randomly shuffled.

[T] Use the remainder estimate and integration by parts to approximate $\sum_{n=1}^{\infty }n/{e}^{n}$ within an error smaller than $0.0001.$

Does $\sum_{n=2}^{\infty }\frac{1}{n{(\ln n)}^{p}}$ converge if $p$ is large enough? If so, for which $p?$

In Exercises 5–12, use the Integral Test to determine the convergence of the given series.

$\displaystyle \sum_{n=1}^\infty \frac{1}{2^n}$

In Exercises 5–12, use the Integral Test to determine the convergence of the given series.

$\displaystyle \sum_{n=1}^\infty \frac{1}{n^4}$

In Exercises 5–12, use the Integral Test to determine the convergence of the given series.

$\displaystyle \sum_{n=1}^\infty \frac{n}{n^2+1}$

In Exercises 5–12, use the Integral Test to determine the convergence of the given series.

$\displaystyle \sum_{n=2}^\infty \frac{1}{n\ln n}$

In Exercises 5–12, use the Integral Test to determine the convergence of the given series.

$\displaystyle \sum_{n=1}^\infty \frac{1}{n^2+1}$

In Exercises 5–12, use the Integral Test to determine the convergence of the given series.

$\displaystyle \sum_{n=2}^\infty \frac{1}{n(\ln n)^2}$

In Exercises 5–12, use the Integral Test to determine the convergence of the given series.

$\displaystyle \sum_{n=1}^\infty \frac{n}{2^n}$

In Exercises 5–12, use the Integral Test to determine the convergence of the given series.

$\displaystyle \sum_{n=1}^\infty \frac{\ln n}{n^3}$

In Exercises 13–22, use the Direct Comparison Test to determine the convergence of the given series; state what series is used for comparison.

$\displaystyle \sum_{n=1}^\infty \frac{1}{n^2+3n-5}$