For the series: `ln(2)/sqrt(2) + ln(3)/sqrt(3)+ ln(4)/sqrt(4)+ ln(5)/sqrt(5)+ ln(6)/sqrt(6) +...`, it follows the formula `sum_(n=2)^oo ln(n)/sqrt(n)` where `a_n = ln(n)/sqrt(n)` . To confirm if the Integral test will be applicable, we let `f(x) = ln(x)/sqrt(x)` .
Graph of the function `f(x)` :
Maximize view:
As shown on the graphs, `f` is positive and continuous on the finite interval `[1,oo)` . To verify if the function will eventually decreases on the given interval, we may consider derivative of the function.
Apply Quotient rule for derivative: `d/dx(u/v) = (u'* v- v'*u)/v^2` .
Let `u = ln(x)` then `u' = 1/x`
`v = sqrt(x)` or `x^(1/2)` then `v' = 1/(2sqrt(x))`
Applying the Quotient rule, we get:
`f'(x) = (1/x*sqrt(x)-1/(2sqrt(x))*ln(x))/(sqrt(x))^2`
`= (1/sqrt(x) - ln(x)/(2sqrt(x)))/x`
`= ((2-ln(x))/sqrt(x))/x`
` =((2-ln(x))/sqrt(x))* 1/x`
`=(2-ln(x))/(xsqrt(x)) `
or `(2-ln(x))/x^(3/2)`
Note that `2-ln(x) lt0` for higher values of x which means ` f'(x) lt0`.
Aside from this, we may verify by solving critical values of x .
Apply First derivative test: f'(c) =0 such that x =c as critical values.
`(2-ln(x))/x^(3/2)=0`
`2-ln(x)=0`
`ln(x) =2`
`x = e^2`
`x~~7.389`
Using `f'(7) ~~0.0015` , it satisfy `f'(x) gt0` therefore the function is increasing on the left side of `x=e^2` .
Using `f'(8) ~~-0.0018` , it satisfy `f'(x) lt0 ` therefore the function is decreasing on the right side of `x=e^2` .
Then, we may conclude that the function `f(x)` is decreasing for an interval `[8,oo)` .
This confirms that the function is ultimately positive, continuous, and decreasing for an interval `[8,oo)` . Therefore, we may apply the Integral test.
Note: Integral test is applicable if f is positive, continuous , and decreasing function on interval `[k, oo)` and `a_n=f(x)` . Then the series `sum_(n=k)^oo a_n ` converges if and only if the improper integral `int_k^oo f(x) dx` converges. If the integral diverges then the series also diverges.
To determine the convergence or divergence of the given series, we may apply improper integral as:
`int_8^oo ln(x)/sqrt(x)dx = lim_(t-gtoo)int_8^tln(x)/sqrt(x)dx`
or `lim_(t-gtoo)int_8^tln(x)/x^(1/2)dx`
To determine the indefinite integral of `int_8^tln(x)/x^(1/2)dx` , we may apply integration by parts: `int u dv = uv - int v du`
`u = ln(x)` then `du = 1/x dx` .
`dv = 1/x^(1/2) dx` then `v= int 1/x^(1/2)dx = 2sqrt(x)`
Note: To determine v, apply Power rule for integration `int x^n dx = x^(n+1)/(n+1).`
`int 1/x^(1/2)dx =int x^(-1/2)dx`
` =x^(-1/2+1)/(-1/2+1)`
`=x^(1/2)/(1/2)`
`=x^(1/2)*2/1`
`=2x^(1/2)` or `2 sqrt(x)`
The integral becomes:
`int_8^t ln(x)/sqrt(x) dx=ln(x) * 2 sqrt(x) - int 2sqrt(x) *1/x dx`
`=2sqrt(x)ln(x) - int 2x^(1/2) *x^(-1) dx`
`=2sqrt(x)ln(x) - int 2x^(-1/2) dx`
`=2sqrt(x)ln(x) - 2int x^(-1/2) dx`
`= [ 2sqrt(x)ln(x)- 2(2sqrt(x))]|_8^t`
`= [2sqrt(x)ln(x) - 4sqrt(x)]|_8^t`
Apply definite integral formula: `F(x)|_a^b = F(b) - F(a)` .
`[2sqrt(x)ln(x) - 4sqrt(x)]|_8^t =[2sqrt(t)ln(t) - 4sqrt(t)] - [2sqrt(8)ln(8) - 4sqrt(8)]`
` =2sqrt(t)ln(t) - 4sqrt(t) - 2sqrt(8)ln(8) + 4sqrt(8)`
` =2sqrt(t)ln(t) - 4sqrt(t) - 4sqrt(2)ln(8) + 8sqrt(2)`
Note: `sqrt(8) = 2sqrt(2)`
Applying `int_8^t ln(x)/sqrt(x) dx=2sqrt(t)ln(t) - 4sqrt(t) - 4sqrt(2)ln(8) + 8sqrt(2)` , we get:
`lim_(t-gtoo)int_2^tln(x)/sqrt(x)dx =lim_(t-gtoo) [2sqrt(t)ln(t) - 4sqrt(t) - 4sqrt(2)ln(8) + 8sqrt(2)]`
`=lim_(t-gtoo) 2sqrt(t)ln(t) - lim_(t-gtoo)4sqrt(t) - lim_(t-gtoo)4sqrt(2)ln(8) + lim_(t-gtoo) 8sqrt(2)`
` = oo-oo -4sqrt(2)ln(8) +8sqrt(2)`
`=oo`
The `lim_(t-gtoo)int_8^tln(x)/sqrt(x)dx=oo` implies that the integral diverges.
Conclusion:
The integral `int_8^ooln(x)/sqrt(x)dx` is divergent therefore the series`sum_(n=2)^ooln(n)/sqrt(n)` must also be divergent.
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