Skip to main content

`sum_(n=2)^oo 1/(n(lnn)^p)` Find the positive values of p for which the series converges.

To find the convergence of the series `sum_(n=2)^oo 1/(n(ln(n))^p) ` where `pgt0` (positive values of `p` ), we may apply integral test.


Integral test is applicable if f is positive, continuous, and decreasing function on an interval and let `a_n=f(x)` . Then the infinite series` sum_(n=1)^oo a_n` converges if and only if the improper integral `int_1^oo f(x) dx ` converges to a real number. If the integral diverges then the series also diverges.


For the infinte series series `sum_(n=2)^oo 1/(n(ln(n))^p)` , we have:


`a_n =1/(n(ln(n))^p)`


Then,` f(x) =1/(x(ln(x))^p).`


The `f(x)` satisfies the conditions for integral test based on the following reasons:


-` f(x)` is continuous since` x(ln(x))^p !=0` for any x-value on the interval `[2,oo)`


-` f(x)` is positive since `1/(x(ln(x))^p)gt0 ` for any x-value on the interval `[2,oo).`


-` f(x)` is decreasing since `f'(x)` is negative  for large value of `x` . It eventually decreases at the tail of the series.


To evaluate the convergence of the series using integral test, we set-up the improper integral as: 


`int_2^oo 1/(x(ln(x))^p)dx`


Apply u-substitution by letting: `u=ln(x)` `u=ln(x)` then `du = 1/xdx` , `a=ln(2)` and `b=oo` .


`int_(ln(2))^oo 1/(x(ln(x))^p) dx=int_(ln(2))^oo 1/(ln(x))^p *1/xdx`


                           `= int_(ln(2))^oo 1/u^pdu`


                          ` =int_(ln(2))^oo u^-p dx`


                          `= u^(-p+1)/(-p+1)|_(ln(2))^oo `  


                          `=u^(-(p-1))/(-p+1)|_(ln(2))^oo`


                           `=1/(u^(p-1)(-p+1))|_(ln(2))^oo`


 Apply definite integral formula: ` F(x)|a^b = F(b)-F(a)` .  


`1/(u^(p-1)(-p+1))|_(ln(2))^oo=1/(oo^(p-1)(-p+1))-1/((ln(2))^(p-1)(-p+1))`


                               `= 1/oo-1/(-p+1)1/(ln(2))^(p-1)`


                              ` =0-1/(-p+1)1/(ln(2))^(p-1)`


                              ` =1/(-p+1)1/(ln(2))^(p-1)`


The integral converges to a real number when `pgt1` .


Thus, the series `sum_(n=2)^oo 1/(n(ln(n))^p)` converges whenever `pgt1` .

Comments

Post a Comment

Popular posts from this blog

Is there a word/phrase for "unperformant"?

As a software engineer, I need to sometimes describe a piece of code as something that lacks performance or was not written with performance in mind. Example: This kind of coding style leads to unmaintainable and unperformant code. Based on my Google searches, this isn't a real word. What is the correct way to describe this? EDIT My usage of "performance" here is in regard to speed and efficiency. For example, the better the performance of code the faster the application runs. My question and example target the negative definition, which is in reference to preventing inefficient coding practices. Answer This kind of coding style leads to unmaintainable and unperformant code. In my opinion, reads more easily as: This coding style leads to unmaintainable and poorly performing code. The key to well-written documentation and reports lies in ease of understanding. Adding poorly understood words such as performant decreases that ease. In addressing the use of such a poorly ...
Post a Comment
Read more

Is 'efficate' a word in English?

I routinely hear the word "efficate" being used. For example, "The most powerful way to efficate a change in the system is to participate." I do not find entries for this word in common English dictionaries, but I do not have an unabridged dictionary. I have checked the OED (I'm not sure if it is considered unabridged), and it has no entry for "efficate". It does have an entry for "efficiate", which is used in the same way. Wordnik has an entry for "efficate" with over 1800 hits, thus providing some evidence for the frequency of use. I personally like the word and find the meaning very clear and obvious when others use it. If it's not currently an "officially documented" word, perhaps its continued use will result in it being better documented.
Post a Comment
Read more