Miscellaneous Problems

Prove Bernoulli's inequality:

where the numbers $\alpha , \beta , \gamma , \ldots , \lambda$ have the same sign and are greater than $-1$.

Using Bernoulli's inequality, prove that for integer $n$, $\left(1-\frac{1}{n^{2}}\right)^{n} > 1-\frac{1}{n}$, if $n > 1$.

Prove that $\left(1+\frac{1}{n}\right)^{n} > \left(1+\frac{1}{n-1}\right)^{n-1}$, if $n > 1$.

Prove that $\left(1+\frac{1}{n}\right)^{n+1} < \left(1+\frac{1}{n-1}\right)^{n}$.

The quantity $\left(1+\frac{1}{n}\right)^{n}$ increases, while the quantity $\left(1+\frac{1}{n}\right)^{n+1}$ decreases, as the integer $n$ increases. From this and from the obvious inequality $\left(1+\frac{1}{n}\right)^{n} < \left(1+\frac{1}{n}\right)^{n+1}$, deduce that both of these quantities tend to a common limit as $n \to \infty$. This limit is denoted by $e$:

Prove the equality:

assuming that the limit on the right-hand side exists.

Prove the equality:

assuming that $\lim_{n \to \infty } \frac{u_{n}}{v_{n}}$ exists and that $v_{1}+v_{2}+v_{3}+\ldots +v_{n} \to \infty$ as $n \to \infty$, with the numbers $v_{n}$ positive.

Prove the equality:

assuming that the limit on the right-hand side exists and $v_{n} \to \infty$ as $n \to \infty$, with $v_{n} > v_{n-1}$.

For positive integer $m$, prove the equality:

Under the same condition, prove the equality:

Prove the equality:

$m$ is an integer $\geq 0$.

Prove the equality:

$m$ is an integer $\geq 0$.

Find $\lim_{n \to \infty } \sum_{k=1}^{n} \left(\sqrt{1+\frac{k}{n^{2}}}-1\right)$.

Find $\lim_{n \to \infty } \sum_{k=1}^{n} \left(\sqrt[3]{1+\frac{k^{2}}{n^{3}}}-1\right)$.

From the inequalities $\left(1+\frac{1}{n}\right)^{n} < e < \left(1+\frac{1}{n}\right)^{n+1}$ it follows that

From this, prove the equality:

Here $C$ is a constant called Euler's constant, and $\varepsilon (n) \to 0$ as $n \to \infty$.

Using the previous result, prove the equality:

Starting from the equality $\lim_{x \to 0} \frac{\ln (1+x)}{x} = 1$, find

Using the formula $\lim_{x \to 0} \frac{\sin x}{x} = 1$, prove that

Prove that as $n \to \infty$

Prove that

Prove the equality

Prove the equality $\lim_{n \to \infty } \left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right) \ldots \left(1-\frac{1}{n(n+1)/2}\right) = \frac{1}{3}$.

Prove the equality $\lim_{n \to \infty } \frac{2^{3}-1}{2^{3}+1} \cdot \frac{3^{3}-1}{3^{3}+1} \cdots \frac{n^{3}-1}{n^{3}+1} = \frac{2}{3}$.

Prove the equality $\lim_{n \to \infty } (1+x)(1+x^{2})(1+x^{4}) \cdots (1+x^{2^{n}}) = \frac{1}{1-x}$; $\left|x\right| < 1$.

Prove the equality $\lim_{n \to \infty } \cos \frac{x}{2} \cdot \cos \frac{x}{4} \cdots \cos \frac{x}{2^{n}} = \frac{\sin x}{x}$ (Euler).

Prove the equality

(Viete).

A sequence of values $u_{n}$ is given by the formulas: $u_{1} = \sqrt{2}$, $u_{2} = \sqrt{2+\sqrt{2}}$, $u_{3} = \sqrt{2+\sqrt{2+\sqrt{2}}}$ and in general $u_{n} = \sqrt{2+u_{n-1}}$. Replacing the last 2 under the radical by 4, verify that $u_{n}$ is bounded. Having proved that $\lim_{n \to \infty } u_{n}$ exists, find it from the equality $u_{n}^{2} = 2+u_{n-1}$.

A sequence $v_{n}$ is given by the equalities:

Replacing the number $a$ under the last radical by the quantity $a+x$, where $x$ is the positive root of the equation $x^{2} = a+x$, verify that $v_{n}$ is bounded, and prove that $v_{n} \to x$ as $n \to \infty$.

A sequence of numbers is given by the equalities

Find an expression for $x_{n}$ in terms of $a$ and $b$, and find $\lim_{n \to \infty } x_{n}$.

Numbers $a$ and $b$ are given, with $a > b > 0$. New pairs of numbers are then formed by the formulas: $a_{n+1} = \frac{a_{n}+b_{n}}{2}$, $b_{n+1} = \frac{2 a_{n} b_{n}}{a_{n}+b_{n}}$. Here $a_{0} = a$ and $b_{0} = b$. Prove that $a_{n} b_{n} = ab$, that

and, finally, that $\lim_{n \to \infty } a_{n} = \lim_{n \to \infty } b_{n} = \sqrt{ab}$.

From two given positive numbers $a$ and $b$ one can construct a series of new numbers using the equalities: $a_{0} = a$, $b_{0} = b$, $a_{1} = \frac{a_{0}+b_{0}}{2}$, $b_{1} = \sqrt{a_{0} b_{0}}$, and in general, $a_{n+1} = \frac{a_{n}+b_{n}}{2}$, $b_{n+1} = \sqrt{a_{n} b_{n}}$. Prove that as $n \to \infty$ the numbers $a_{n}$ and $b_{n}$ tend to a common limit. (Gauss's arithmetic-geometric mean.)

Two sequences of numbers are formed by the formulas:

Here $a_{0} = a$, $b_{0} = b$ are given positive numbers. Find $\lim_{n \to \infty } a_{n}$, which equals $\lim_{n \to \infty } b_{n}$, setting $a = b \cos \varphi$ for $a < b$ and $a = b \operatorname {ch} \varphi$ for $a > b$.

A sequence of numbers is formed by the formulas: $x_{n} = \frac{1}{2}\left(x_{n-1}+\frac{a}{x_{n-1}}\right)$, where $a > 0$ is a given number and $x_{0}$ is an arbitrary positive number. Prove that $x_{n} \to \sqrt{a}$ as $n \to \infty$.

The numbers of a sequence are formed according to the following law: $x_{0}$ is an arbitrary positive number, and the subsequent ones are found by the formula

where $a$ is a given positive number. Prove that $x_{n} \to \sqrt{a}$ as $n \to \infty$.

Prove that the sequences formed, similarly to the earlier ones, by the formula $x_{n+1} = \frac{1}{3}\left(2x_{n}+\frac{a}{x_{n}^{2}}\right)$ or by the formula $x_{n+1} = \frac{x_{n}(x_{n}^{3}+2a)}{2x_{n}^{3}+a}$, tend to $\sqrt[3]{a}$.

A sequence $y_{n}$ is defined as follows: $y_{1} = \frac{x}{2}$, where $0 < x < 1$; $y_{2} = \frac{x}{2}+\frac{y_{1}^{2}}{2}$, and in general $y_{n} = \frac{x}{2}+\frac{y_{n-1}^{2}}{2}$. Prove that $y_{n}$ is a bounded monotone quantity, and find $\lim_{n \to \infty } y_{n}$.

$y_{1} = \frac{x}{2}$, where $0 < x < 1$; $y_{2} = \frac{x}{2}-\frac{y_{1}^{2}}{2}$; ; $y_{n} = \frac{x}{2}-\frac{y_{n-1}^{2}}{2}$. Show that $\lim_{n \to \infty } y_{n}$ exists and equals $\sqrt{1+x}-1$.