1/sqrt (n + 1) convergent or divergent?

Question

What method should I use to solve this? Infinity Summation n =1. I was thinking p-series but I don't think I can break up the sqrt n + 1 .

Answer 1

You can still use p-series. You can just change the indices on summation:



∑[n = 0..∞] 1/√(n+1)



Let n+1 = i

When n = 0, i = 1



∑[n = 0..∞] 1/√(n+1) = ∑[i = 1..∞] 1/√i



By p-series ∑[i = 1..∞] 1/√i diverges, since 0 < p ≤ 1

Therefore ∑[n = 0..∞] 1/√(n+1) diverges, since it is = ∑[i = 1..∞] 1/√i



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Alternate method: p-series combined with comparison test:



For n ≥ 1

n + 1 ≤ 2n

√(n+1) ≤ √(2n)

1/√(n+1) ≥ 1/√(2n)



By p-series ∑[1..∞] 1/√(2n) = 1/√2 ∑[1..∞] 1/n^(1/2) diverges, since p = 1/2 ≤ 1



By comparison test, ∑[1..∞] 1/√(n+1) also diverges



If series actually starts from 0, then we can show that it also diverges, since the sum of a finite value and a diverging series is also diverging:

∑[0..∞] 1/√(n+1) = 1 + ∑[1..∞] 1/√(n+1)



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Alternate method: Integral test:



∫[0 to ∞] 1/√(x+1) dx

= 2 √(x+1) |[0 to ∞]

= 2 { (lim[x→∞] √(x+1)) - √1 }

= 2 (∞ - 1)

= ∞



----> integral diverges



By integral test, series diverges



Mαthmφm

Answer 2

This is divergent. We know this since

∑1/(n+1)

is divergent since it is the harmonic series and

1/√(n+1) > 1/(n+1) for all n >= 0

Answer 3

if n is helpful,n=a million,2,3,4,..... the respond is divergent,as a million divides with the aid of sqrt(n) turns into smaller n smaller until it reaches 0 if n is unfavourable,then the equation is undefined.