# Practice Math With Us # P Series Calculus

Introduction:

In mathematics we will call the harmonic series as the divergent infinite series. If an infinite series is having the variable we will call that is p series calculus function. Normally p series calculus is given by the closed form of Riemann zeta function. Let us see the following function.

`sum_(k=1)^oo k^(-p) = zeta(p)`

Herep is the parameter. If p > 0. So we are using the p series calculus function. If we sum the p series we are summing the smaller values. Thisis better example to do the Converges.

`sum_(k=1)^oo (1 / k^p)`

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## Explanation for P series calculus:

P series calculus function is nothing but the power series. In p seriesfunction the main term is k. if any function contains the k value and the power is p series than we can say it is p series function.

Let us see an example for p series,

`sum_(k=1)^oo(1 / k^p)`

We can do the following test using the above p series,

If p > 1 the given series   `sum_(k=1)^oo(1 / k^p)`

Else if p `<=1`  ` sum_(k=1)^oo(1 / k^p) = oo`

Let us see some example for p series and converges test.

## Test based on p series calculus:

Example for p series:

1. `sum_(k=1)^oo = (1 / k^2)`

2. `sum_(k=1)^oo = (1 / k!)`

Nowwe will see some conditions to do the divergence test, comparison test,Limit comparison test, Ratio test, Root test and alternating testing series.

Divergence test:

Let us assume the following p series

`sum_(k=1)^oo = (1 / k^p)`

If p > 1 than it is diverges. `0 < p <= 1`

Comparison test:

Let us take any two p series x k and y k. than `"sum `x k and y k

Where x k < y k is large

I) If `sum `x k is converges than `sum `y k is also converges.

II) If `sum `x k is diverges than `sum `y k is also diverges.

Limit comparison test:

If `sum `x k and `sum `y k is having positive values than `lim_(k->oo) (`(x k / y k) = L

Where `0 < L <= oo`

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Ratio Test:

If `sum `x k is having positive values than `lim_(k->oo) (`(x k + 1 / x k) = L

1. Where L < 1 `sum `x k is converges

2. L > 1 than `sum `x k is diverges

3. L = 1 than `sum `x k is inconclusive test.

Root test:

If `sum `x k is having non - negative values than `lim_(k->oo) (`(x k)1 / k= L

1. Where L < 1 `sum `x k is converges

2. L > 1 than `sum `x k is diverges

3. L = 1 than `sum `x k is inconclusive test.

Alternative test series:

In the alternative series `sum_(k=1)^oo (-1)^k` x k

Here x k > 0 where all  `k >= 0`