Determine if series is con/divergent

[bfusco]

Homework Statement

Ʃ[√(n+1)-√(n)]/√(n^2 +n)

The Attempt at a Solution

well i split the series up first.

Ʃ√(n+1)/√(n^2 +n) - Ʃ√(n)/√(n^2 +n)

next:

Ʃ[(n+1)/(n^2 +n)]^1/2 - Ʃ[n/(n^2 +n)]^1/2

at this point I am not sure if i can take the limit and divide all the terms by the highest power of n.

 

[Dick]

Both of the series you've split it up into diverge. They are like 1/sqrt(n). I'd suggest you multiply numerator and denominator by sqrt(n+1)+sqrt(n) first. It's the quadratic conjugate trick. Then use a comparison test.

 

[bfusco]

Homework Statement

Ʃ[√(n+1)-√(n)]/√(n^2 +n)

The Attempt at a Solution

well i split the series up first.

Ʃ√(n+1)/√(n^2 +n) - Ʃ√(n)/√(n^2 +n)

next:

Ʃ[(n+1)/(n^2 +n)]^1/2 - Ʃ[n/(n^2 +n)]^1/2

at this point I am not sure if i can take the limit and divide all the terms by the highest power of n.

so I am basically using the comparison test? I am comparing these series to 1/n, which is divergent, and because these series are less than 1/n it also is divergent. also considering that 1 of the now 2 series is diveregent the whole thing is divergent.

 

[Dick]

so I am basically using the comparison test? I am comparing these series to 1/n, which is divergent, and because these series are less than 1/n it also is divergent. also considering that 1 of the now 2 series is diveregent the whole thing is divergent.

The two series you've split it up into diverge. The original series does not diverge. Read my post again.

 

[DryRun]

If you split up the following fraction into its sum;
\sum^{\infty}_{n= ?}\frac {\sqrt{n+1}-\sqrt n}{\sqrt {n^2 +n}}=\sum^{\infty}_{n= ?}\frac {\sqrt{n+1}}{\sqrt {n^2 +n}}-\sum^{\infty}_{n= ?}\frac {\sqrt n}{\sqrt {n^2 +n}}
and then take the limits for both, you will get zero in each case, which by the nth-term test for divergence, doesn't say if both series converge or diverge.

Both of the series you've split it up into diverge. They are like 1/sqrt(n). I'd suggest you multiply numerator and denominator by sqrt(n+1)+sqrt(n) first. It's the quadratic conjugate trick. Then use a comparison test.

Normally, isn't the multiplication of the conjugate done with the denominator, instead of the numerator as you suggested?

 

[LCKurtz]

Normally, isn't the multiplication of the conjugate done with the denominator, instead of the numerator as you suggested?

There isn't any "normally". You can rationalize either a numerator or denominator. What you do depends on your problem and what your goal with the particular problem is. Dick's post is spot on for a good suggestion.