Series: I really need help with these?

Q1.

Let a(n) = n(x - 4)^n / (n^4 + 1)

Thus, a(n + 1) = (n + 1)(x - 4)^(n + 1) / ((n + 1)^4 + 1)

limit of |a(n + 1) / a(n)| as n tends to infinity

= lim | [(n + 1)(x - 4)^(n + 1) / ((n + 1)^4 + 1)] / [n(x - 4)^n / (n^4 + 1)] |

= lim |(n + 1) / n| |(x - 4)| |((n^4 + 1) / ((n + 1)^4 + 1)|

= lim |1 + (1/n)| lim |x - 4| lim |((n^4 + 1) / ((n + 1)^4 + 1)||

= (1) |x - 4| (1)

= |x - 4|

By the ratio theorem, |x - 4| must be < 1 for the series to be convergent.

|x - 4| < 1

-1 < x - 4 < 1

3 < x < 5

Thus, the radius of convergence is just 1.

At x = 5, the summation becomes n / (n^4 + 1)

Notice that the terms of this series are all non-negative.

Also, n / n^4 = 1 / n^3 > n / (n^4 + 1)

The summation of 1 / n^3 from 1 to infinity is a p-series with p = 3.

Hence, it is convergent.

By the comparison test, summation of n / (n^4 + 1) is convergent.

At x = 3, the summation becomes (-1)^n (n) / (n^4 + 1)

Taking absolute values, we have the summation of n / (n^4 + 1).

We have already found out that this is convergent.

Thus, by theorem, (-1)^n (n) / (n^4 + 1) is absolutely convergent,

which makes it convergent as well.

Hence, the interval of convergence is 3 <= x <= 5

Q2.

Let a(n) = (-1)^n x^n / (4^n ln n)

Thus, a(n + 1) = (-1)^(n + 1) x^(n + 1) / (4^(n + 1) ln (n + 1))

limit of |a(n + 1) / a(n)| as n tends to infinity

= lim |[ (-1)^(n + 1) x^(n + 1) / (4^(n + 1) ln (n + 1))] / [(-1)^n x^n / (4^n ln n)]

= lim |x| |1/4| |ln n / ln (n + 1)|

= (1/4)|x| lim |ln n / ln (n + 1)||

= (1/4)|x| (1)

= (1/4) |x|

By the ratio test, the summation is convergent only if (1/4) |x| < 1

|x| < 4

Thus, the radius of convergence is 4.

When x = -4, the summation becomes:

(-1)^n (-4)^n / (4^n ln n)

= 4^n / (4^n ln n)

= 1 / ln n

n > ln n for all positive integer n, n >= 2, so

1/n < 1/ln n for all positive integer n, n >=2.

Thus, the summation of 1 / n is less than the summation of 1 / ln n

from 2 to infinity.

Note that the terms of both series are non-negative.

The summation of 1 / n is the harmonic series, which is divergent.

By the comparison test, the series of 1 / ln n is also divergent.

When x = 4, the summation becomes:

(-1)^n 4^n / (4^n ln n)

= (-1)^n / ln n

We got an alternating series.

Let b(n) = 1/ ln n

Thus, b(n + 1) = 1 / ln (n + 1)

limit of b(n) as n tends to infinity is zero.

Notice that b(n) is non-negative for all positive integer n >= 2.

ln n < ln (n + 1) for all positive integer n >=2

1 / ln n > 1 / ln (n + 1) for all positive integer n >= 2

Thus, b(n) > b(n + 1) for all positive integer n >= 2

Hence by the alternating series, the series is convergent when x = 4

This means that the interval of convergence is -4 < x <= 4

Q3a)

Let a(n) = (2x)^n / (2n)!

Thus, a(n + 1) = (2x)^(n + 1) / (2n + 2)!

limit of |a(n + 1) / a(n)| as n tends to infinity

= lim |[(2x)^(n + 1) / (2n + 2)!] / [(2x)^n / (2n)!]|

= lim |(2x) / (2n + 1)(2n + 2)|

= |2x| lim |1 / (2n + 1)(2n + 2)|

= |2x| (0)

= 0

By the ratio test, the summation converges only if 0 < 1

But clearly 0 is < 1 for all real numbers.

Hence, the interval of convergence is: - infinite < x < infinite

Q3b)

For the previous three questions, we used the ratio test.

Now, we will use the root test instead.

Let a(n) = ((x - 1) / ln n)^n

By the root test, the series is convergent only if

limit of (nth root of |a(n)|) as n tends to infinity is < 1

limit of (nth root of |a(n)|) as n tends to infinity

= lim |(x - 1) / ln n|

= |(x -1)| lim |1 / ln n|

= |x - 1| (0)

= 0

0 < 1 for all real x

Thus, interval of convergence is once again:

-infinity < x < infinity