**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)`

Please express your views of this topic Dot Product Properties by commenting on blog.

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.

**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`

Between, if you have problem on these topics Derivative of a Log Function,please browse expert math related websites for more help on Derivative of a Natural Log .

**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`