Judge the convergence and divergence of series. ∑ (n = 1 →∞) (radical n + 1 minus radical n)

Judge the convergence and divergence of series. ∑ (n = 1 →∞) (radical n + 1 minus radical n)

sn=√2-√1+√3-√2+√4-√3+.+√n+1-√n=√n+1 -1
The limit is + ∞
therefore
Divergence

Why does n + 1-radical n not converge

Since 1 / (√ (n + 1) + √ n) > 1 / (2 √ (n + 1)) > 1 / (2 (n + 1)) > 1 / (2 (n + 1)), and the series ∑ 1 / (2 (n + 1)) diverges, the series ∑ (√ (n + 1) - √ n) diverges

Does the series ((- 1) ^ [radical n]) / N converge

Judge whether the series ∑ (n = 1) (- 1) ^ n / (n + radical n) is absolute convergence, conditional convergence or divergence

The {an} is a Leibniz staggered series, so it converges
1 / (n + radical n) > 1 / (n + n) = 1 / 2n, because {1 / 2n} diverges, so {an │} also diverges
Therefore, the {an} condition converges

Is the series (- 1) ^ n multiplied by ln (1-1 / radical n) convergent or divergent

This is an alternating series, which is known to converge by Leibniz theorem. The economic mathematics team will help you solve it, please evaluate it in time

Judge the convergence and divergence of alternating series: sin (square of N under π × root sign + 1) from n = 1 to infinity

Convergence, as shown in the figure

Judge the convergence and divergence of n (N2 + 1) under + ∞Σ n = 1 1 / radical

A positive series of 1 / N ^ P level converges as long as P is strictly greater than 1, and diverges as long as P is equal to or less than 1

Discriminant radical 2 + radical 3 / 2 + +Convergence of radical ((n + 1) / N)

(n + 1) / N is always greater than 1, so you can imagine that its image should be above y = x, so it is impossible to converge. As long as the limit of a polynomial whose term is positive is not 0 when n reaches positive infinity, it is impossible to converge

If f (x) = 2 x power (denominator) minus 1 / 1 + A is an odd function, then a=____

It's like - 1

Lim n →∞, where X1 = 1, xn + 1 = √ (2xn + 3), n > = 1

Prove that xn is bounded
Conjecture: xnxn
Then, xn increases monotonically
Because xn is monotonically increasing and bounded, according to the monotone bounded theorem:
Xn convergence
Let Lim xn = a
Because: X (n + 1) = √ (2xn + 3)
Take the limit at the same time
lim x(n+1)=lim √(2xn+3)
a=√(2a+3)
A = 3 or a = - 1 (rounding off)
Therefore,
lim xn=3
If you don't understand, please ask