I need help on calculus. proves, limits and find sum

[mild steeple]

What way should i solve A

In A i need to find the sum

[covert vector]

Firstly, notice that \begin{equation} \sum_{n=1}^{N}f(n)-f(n+k)=\sum_{n=1}^{k}f(n)-f(n+N) \end{equation}

[fast fulcrum]

otiopiscis

[covert vector]

Since [\frac{-4}{n^2+5n+4}=\frac{-4}{3}\left(\frac{1}{n+1}-\frac{1}{n+4}\right) ],

[mild steeple]

The - is on the other side for me

Like -1/n+1

[covert vector]

Emm... maybe it is not a very serious problem, definitely i can write the whole solution done

[fast fulcrum]

otiopiscis

[mild steeple]

Why did you add - before the 4

[covert vector]

there is a small "-" in your picture

[mild steeple]

Hm... ok

Can you show me how do you break the 4/n^2+5n+4 ?

[covert vector]

Sure, $n^2+5n+4=(n+1)(n+4)$

[fast fulcrum]

otiopiscis

[covert vector]

So that $$\frac{4}{(n+1)(n+4)}$$ is $$\frac{4}{3}\left(\frac{1}{n+1}-\frac{1}{n+4}\right) $$

[fast fulcrum]

otiopiscis

[mild steeple]

Ok

[covert vector]

Well, then we can apply formula $(1)$ to the sum of it,

[fast fulcrum]

otiopiscis

[mild steeple]

And then how can i open it? Like this?

[covert vector]

Yeah

[mild steeple]

Ok thx.
For B. I need to check first if it convergents or not

And find sum if is

[covert vector]

Yeah ur correct. After checking, simply using the relation $\ln\frac{n+1}{n}=\ln{(n+1)}-\ln{n}$ and using $(1)$ again 🙂

[fast fulcrum]

otiopiscis

[mild steeple]

How do i show first if it convergents before i show the sum

With the lim an+1/an?

[covert vector]

🤔 I think that one of the most efficient way to show the convergence (in this problem) is to find the result of the finite sum at first, and then find the limit of it

Ooh the infinity sum in B doesn't convergent, the result is infinity

[mild steeple]

Yeah

@covert vector i have another question not on this problem. Lets say i have proven that my problem is convergents. Does it mean the limit of the general statement is 0?

[covert vector]

yeah, a general result it that "the n-th term of the series converges to 0" is just a necessary-but-not-sufficient condition of the statement "the series is convergent"; if u have known that some series (infinity sum) is convergent then u can imply that the n-th term of the series converges to 0, but vise versa is not.

[mild steeple]

I mean like that

[covert vector]

Yeah Your statement is correct )

[mild steeple]

Ok thx lad

Really appreciate it

[covert vector]

U r welcome 🙂

[mild steeple]

I have so much work to do on that. And will

My questions will pop here

[modest field]

I need help on calculus. proves, limits and find sum

[mild steeple]

same here right?

[covert vector]

Well, aalsoo right here. We can use some specific approaches to find the result of some limits with this form, but it may needs a lot minutes for me to find the correct method 🤔 ...

One possible method is to compare the limit with some easy guys. We can prove that $n!>{\left(\frac{n}{e}\right)}^n$ at first, then the next steps would be trivial

[fast fulcrum]

otiopiscis

[covert vector]

To prove it, let $$x_n={\left(\frac{n}{e}\right)}^n,$$ then we can find that $$x_n=x_1\frac{x_2}{x_1}\frac{x_3}{x_2}\ldots\frac{x_n}{x_{n-1}}<n!$$ since $$\frac{x_n}{x_{n-1}}=\frac{n{\left(1+\frac{1}{n-1}\right)}^{n-1}}{e}<n$$

[fast fulcrum]

otiopiscis

[strong bramble]

.close