alternating series-are these correct?

Question

Can someone tell me if these are right? I hate it how textbooks only give answers to odd exercises.

Alternating series test:
oo
[sigma] [(-1)^(n+1)]n/2n-1
n=2

diverges because n-th term test fails

oo
[sigma] (-1)^n/ln(n+1)
n=1
diverges because the n-th term test fails

oo
[sigma] [(-1)^n+1]n/n^2+1
n=1

diverges because the second test fails

are these right??
thanks!

Answer 1

have another think about the last

it should not be far from alternating 1/n
take terms in pairs and see what you get

n/(n^2+1) - (n+1)/((n+1)^2+1)
= 1/(n+1/n) - 1/( n+1+1/(n+1))
= ( n + 1 + 1/(n+1) - (n+1/n) ) / [(n+1/n)(n+1+1/(n+1))]
= ( 1 -1/(n(n+1)) ) / [(n+1/n)(n+1+1/(n+1))] = O(1/n^2)

Answer 2

The necessary(by no means sufficient) condition for the convergence of any kind of series is that the nth term has limit ZERO
So,if the limit is not 0 it does not matter if the series is alternate.It´s enough to say that series is divergent
The first is divergent because lim Ia_nI is not 0
the second converges by Leibnitz
Let´s take up the last
a_n = (-1)^n *n/(n^2+1) + n/(n^2+1)
The secon term is not alternate and is the nths term of a series of the same kind as 1/n which diverges to +infinity but this is not enough because the first term can belong to a series divergent to - infinity.So you have to study this series

a_n= (-1)^n n/(n^2+1)

lim a_n= 0 2) lets see if it is monotonos decreasing in absolute value.If you take Ia_(n+1)I-I a_nI it turns out to be negative.So it is decreasing and by Leibnitz is convergent
Just now you can say that your given series is divergent.

Answer 3

Only the first one is right. The second two are both conditionally convergent.

For the first one, we indeed have that [n=2, ∞]∑((-1)^(n+1)n/(2n-1)) diverges, because [n→∞]lim ((-1)^(n+1)n/(2n-1)) does not exist ([n→∞]lim |((-1)^(n+1)n/(2n-1)| = 1/2, but the series keeps flipping between values close to 1/2 and values close to -1/2). Note that even if we removed the (-1)^(n+1), the series would still diverge, because although the limit of the terms would exist, it would not be equal to zero.

[n=1, ∞]∑((-1)^n/ln (n+1)) _converges_, because it is an alternating series, and [n→∞]lim 1/ln (n+1) = 0. The natural logarithm does indeed grow infinitely large -- it does so slowly, but it does so.

Similarly, [n=1, ∞]∑((-1)^(n+1)n/(n²+1)) converges, because it is an alternating series, and [n→∞]lim n/(n²+1) = 0. (Note that I'm assuming you actually meant [n=1, ∞]∑((-1)^(n+1)n/(n²+1)) and not [n=1, ∞]∑(((-1)^n+1)n/(n²+1)) = [n=1, ∞]∑((-1)^n n/(n²+1) + n/(n²+1)), which is divergent).

Answer 4

They're all correct except the last one, which converges by the alternating series test.
The signs of the terms alternate, and the absolute value of the terms decrease.
To see that the absolute value decreases as n increases, write it as
n / (n^2 + 1) = 1 / (n + 1/n)
Then it's easy to see that this denominator increases when n increases by 1.

Answer 5

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Answer 6

First diverges, the second converges, the third converges.