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Where does the series log( 1 + x*x / n*log(n)*log(n) ) converge ?

(x is real)

(x is real)

converges for all x

hint:

log(1+x)<=x

2^k*a(2^k)

hint:

log(1+x)<=x

2^k*a(2^k)

Is it true that when a power series with center 0 converges for
z0, then it converges absolutely for |z|<|z0| ?

Yes. This proof is very easy.

Suppose that the limit |a[n]| / |a[n+1]| exists. Show that it
is the radius of convergence of the power series a[n]*z^n.

Use the ratio test for convergence of series.

Prove that 1 / lim_sup |a[n]| ^ (1/n) is the radius of
convergence of the power series a[n]*z^n.

Use the Cauchy test for convergence of series.

Find the radius of convergence of the series

x^n / ( 2^n + 3^n + (-1)^n * (3^n - 2^n) ).

x^n / ( 2^n + 3^n + (-1)^n * (3^n - 2^n) ).

R = 2

Use Cauchy test.

Use Cauchy test.

Find the radius of convergence of the series ( n! / 2^(n*n) ) *
x^n.

R = oo.

the ratio test

the ratio test

Let p[n] be the sequence of consecutive prime numbers. Find the
radius of convergence of the power series x^p[n] / p[n].

R = 1

Notice that the disputed series converges for -1.

Hence the radius must be at least 1.

Notice that if a>1, then the series diverges for a.

Hence the radius cannot be greater than 1.

Conclusion: R=1

Notice that the disputed series converges for -1.

Hence the radius must be at least 1.

Notice that if a>1, then the series diverges for a.

Hence the radius cannot be greater than 1.

Conclusion: R=1

Let p[n] be the sequence of consecutive prime numbers. Find the
radius of convergence of the power series p[n] * x^p[n].

R = 1

way1:

a[n] = { n for prime n; 0 otherwise }

Use the Cauchy test.

way2:

0<x<1 => +(n=1 to oo) p[n]*x^p[n] < +(n=1 to oo) n*x^n < oo

way1:

a[n] = { n for prime n; 0 otherwise }

Use the Cauchy test.

way2:

0<x<1 => +(n=1 to oo) p[n]*x^p[n] < +(n=1 to oo) n*x^n < oo

Where does this series converge?

1/(x^n) * sin(1/(2^n))

1/(x^n) * sin(1/(2^n))

|x| > 1/2

Hint: lim sin(x)/x = 1 as x approaches 0.

Hint: lim sin(x)/x = 1 as x approaches 0.

By referring to the Euclidean two-dimensional space, justify
intuitively why

[A x B = 0] <=> [A,B are perpendicular].

A x B is the scalar product (a,b)x(c,d) = (a*c,b*d)

[A x B = 0] <=> [A,B are perpendicular].

A x B is the scalar product (a,b)x(c,d) = (a*c,b*d)

Hint: |A-B| = |A+B|