Need math help??

Question

a.Prove that for any positive x<1,the series 1-x+x^2-x^3+.....+x^n coverages to 1/(1-x) as n tends to infinity.
b.Harmonic series 1+1/2+1/3+1/4+....+1/n tends to infinity wit n.

correction:
by harmonic series the sum 1+1/2+1/3+1/4+....+1/n tends to infinity wit n

Answer 1

The first is a geometric series with r=-x which converges to
S= 1/(1+x) as lon as IxI<1
You shoul know that the sum of the n terms of a geometric
is
Sn=r^n-1 *r -1/(r-1)and if IrI<1 lim r^n=0
2) You can take the function y=ln x and apply the mean value theorem to the interval n,n+1
so ln(n+1)-ln(n )= 1/(n+a) where 0 so
ln(n+1)-ln(n)<1/n
Sum up both sides from n=1 t n=n
ln2-ln1<1
ln3-ln2<1/2
Finally on the first side only remains
ln(n+1) < 1+/2+1/3+++1/n
as ln(n+1) has limit + infinity ans the nth sum of the harmonic is > it has limit + infinity

Answer 2

[EDIT: If this is for a calculus class, use the answer above. By the word "prove" in your question, I assumed this was for an analysis class in which the validity of the integral may or may not have been proven yet.]

I can give you the general outline.

a) You're proving convergence, so use the standard opener (i.e. "Let epsilon > 0 be given"). Now go back to the definition: you need to find some N such that, if n>N, then the summation from 1 to n differs from 1/(1-x) by less than epsilon. N can depend on x and epsilon.

b) There are several ways to do this. On the one hand, you could use the integral test or the p-test and be done in one minute. On the other hand, I doubt that's what the exercise intended. Here's a better way (this is just a sketch, the rigor is your job!). The first term is greater than 1/2. The second term is equal to 1/2. The sum of the third and fourth terms is greater than 1/2. The sum of terms 5 through 8 is greater than 1/2. The sum of terms 9 through 16 is greater than 1/2. And so on. Suppose there is some finite limit S, and argue by contradiction: show that S is not a limit.

Answer 3

From algebra we know that the sum of the 1st n terms of a geometric progression is
S= (1-r^n)/(1-r) = 1/(1-r) -r^n(1-r)
If |r| <1, that is if -1 inf.
So S= 1/(1-r) or in your case 1/(1-x).

Note tha x can be positive or negative so it woorks for alternating series as well

The harmonice series can be written 1 +1/2 +(1/3 +1/4)+
(1/5+1/6+1/7+1/8) +(1/9 ,...1/16) +(117+...132)...
taking twice as many terms in each group. The sum of the terms in eac parentheses is >1/2. Therefore the harmonic series is divergent.

Answer 4

1/(1-x) = 1+x+x^2+x^3......................
1/(1+x)= 1-x+x^2-x^3.....................

(In your Qn you have wrongly mentioned the above)

1+x+x^2+x^3......................
Coventionally by considering the given series as Geometric series , the sum of n terms is a(1-r^n)/(1-r) : here a=1 & n terds to infinity & r=x , when x <1, x^n will tend to zero & hence sum will be 1/(1-x).

This can be proved as follows also ( The proof given by Indian Mathematician Nilakanta( AD 1444-1544) ,who for the first time did this infinte sum.

1/(1-x) = (1-x+x)/ (1-x) = 1+ x/(1-x) = 1+ x{ 1 + x/(1-x)}

= 1+x+ x^2/(1-x) = 1+x+x^2( 1+ x/(1-x))= 1+x+x^2+x^3/(1-x)
Thus if it is continued we are getting the series expansion.

The other part of the question, sum of reciprocals is a well known divergent series.

Answer 5

see others solution..lol