Question

Qno.1) Establish the convergence or divergence of the sequence (Xn) such that
I) Xn= 1/1^2+1/2^2+1/3^2+.......1/n^2 for all n belongs to N.
ii) Xn= (1+1/n)^n+1 iii) Xn= (1+1/n+1)^n
Iv) Xn= (1-1/n)^n

Answer 1

Answer on Question #82677 – Math – Real Analysis

Question

Establish the convergence or divergence of the sequence $(X_n)$ such that

i) $X_n = 1/1^2 + 1/2^2 + 1/3^2 + \dots + 1/n^2$ for all $n$ belongs to $N$.

ii) $X_n = (1 + 1/n)^n + 1$

iii) $X_n = (1 + 1/n + 1)^n$

iv) $X_n = (1 - 1/n)^n$

Solution

1) $x_n = \frac{1}{1^2} + \frac{1}{2^2} + \dots + \frac{1}{n^2}$

It is obvious that this sequence is monotonous. Let us prove that it is bounded.

$$\begin{array}{l} x_n < 1 + \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{(n - 1) \cdot n} = 1 + 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \dots + \frac{1}{n - 1} - \frac{1}{n} \\ = 2 - \frac{1}{n} < 2 \end{array}$$

Sequence is monotonous and bounded, and therefore converges.

2a) $x_n = \left(1 + \frac{1}{n}\right)^n + 1$,

$\lim_{n \to +\infty} \left(1 + \frac{1}{n}\right)^n + 1 = e + 1$. Hence, it converges.

2b) $x_n = \left(1 + \frac{1}{n}\right)^{n + 1}$,

$\lim_{n \to +\infty} \left(1 + \frac{1}{n}\right)^{n + 1} = \lim_{n \to +\infty} \left(1 + \frac{1}{n}\right)^n \left(1 + \frac{1}{n}\right) = e \cdot 1 = e$, hence it converges.

3a) $x_n = \left(1 + \frac{1}{n} + 1\right)^n$

$$x_n = \left(2 + \frac{1}{n}\right)^n > 2^n$$

$x_n$ does not converge, or, in other words, converges to $+\infty$.

3b) $x_n = \left(1 + \frac{1}{n + 1}\right)^n$

$$\lim_{n \to +\infty} \left(1 + \frac{1}{n + 1}\right)^n = \lim_{n \to +\infty} \frac{\left(1 + \frac{1}{n + 1}\right)^{n + 1}}{1 + \frac{1}{n + 1}} = \frac{e}{1} = e$$

It converges.

4) $x_n = \left(1 - \frac{1}{n}\right)^n$

$$\left(1 - \frac{1}{n}\right)^n = \left(\frac{n - 1}{n - 1 + 1}\right)^n = \frac{1}{\left(1 + \frac{1}{n - 1}\right)^n} = \frac{1}{\left(1 + \frac{1}{n - 1}\right)^{n - 1} \left(1 + \frac{1}{n - 1}\right)},$$

hence

$$\lim _ {n \to + \infty} x _ {n} = \frac {1}{e ^ {- 1}} = \frac {1}{e ^ {-}}.$$

Thus, $x_{n}$ converges.

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