# My way of solving infinite series

To understand this concept you must understand a variable number, a variable number is a number dedicated to or denotes a variable in a series, this number can have more than one variable but can't share a variable with other numbers, it also can't be a decimal. Ex1:

$\sum_{x=1}^∞ a \frac{b}{c}$

In the example x=1, right? 1 is the variable number and X is its dedicated variable(seems simple but trust me, it can be any number that's not ∞ or a decimal) this is what most of these problems look like:

$\sum_{x=5}^∞ 2\frac{4\times x^{2}}{10} = \frac{a}{b}$ Find a+b

$x=5$ $x^{2}=25$ $25\times 4 =100$ $\frac{100}{10}=100+10=\boxed{110}$

As you saw converging is not done in this method,the way around this is a dummy answer:

$\frac{100}{10}=\boxed{10}$

I hope you consider this a lesson for my future problems, take care!

Note by Ark3 Graptor
2 years, 11 months ago

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I am afraid you are missing the point about what a series is. All your method seems to be doing is finding the first term in a sum, and not the sum itself.

By putting $$x=1$$ in the expression $\sum_{x=1}^\infty 2^{-x}$ your method gets $$\tfrac12$$, and not the correct answer of $$1$$. If (as you do) you only consider the first term of a series, you will continue to get finite answers to series that have no finite limit.

Incidentally, your notion of a variable number is what is normally called a dummy variable. It is a place-holder variable used to indicate how to perform summations, integrals or other calculations. For example. in $\sum_{x=1}^\infty \frac{1}{x^2}$ $$x$$ is an integer-valued dummy variable which can take any value greater than $$1$$, while in $\int_0^3 \sin x\,dx$ $$x$$ is a real-valued dummy variable which can take any value between $$0$$ and $$3$$. For that matter (just to show that dummy variables are not confined to sums and integrals), in $\bigcup_{j=1}^5 A_j \; = \; A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5$ $$j$$ is an integer-valued dummy variable taking values between $$1$$ and $$5$$.

- 2 years, 11 months ago

Thank you for your input, but one thing that bugged me in your comment was the part where the answer was 1/2 and not 1, that is where an error was, you have to not treat it like a fraction if its not a fraction, for example: $\sum_{x=1}^\infty 4\times x^{2}$ There are the three formulas and we have to pick the correct one $\sum_{x=1}^\infty 4\times x^{2}=a,a\frac{b}{c} or \frac{a}{b}$

The correct formula for this series is the first formula

The rest made sense, thanks again

- 2 years, 11 months ago

I am sorry, but this is nonsense. There is no correct formula for the series you mention, which diverges.

Your method would pick $$x=1$$ and get the answer of $$4$$ here, I suppose. The whole point of a dummy variable here is that it indicates the range of values over which the sum is to be performed. Thus, as well as $$x=1$$ giving $$4$$, we have to add the contribution from $$x=2$$, which is $$16$$, then the contribution from $$x=3$$, which is $$36$$ and so on, including the contribution from all positive values of $$x$$. The expression $\sum_{x=1}^\infty 4x^2$ is mathematical shorthand for the sum $4 + 16 + 36 + 64 + 100 + 144 + \cdots$ which does not converge to any value.

- 2 years, 11 months ago

I'm going to have to think about this and do some testing, I will decide later if you are right or I have more of an argument

- 2 years, 11 months ago

Sorry,That was my bad excision of certain things

- 2 years, 11 months ago

I am not quite sure what you mean by that.

I am keen that you take some time to try to understand convergence properly. Isham Singh has given you a link to a suitable Wiki, which would be a good point to start.

- 2 years, 11 months ago

$\sum_{a=1}^{\infty}\sum_{b=2}^{\infty}\sum_{c=3}^{\infty} \dots \sum_{x=24}^{\infty}\sum_{y=25}^{\infty}\sum_{z=26}^{\infty} \dfrac{a+b^2+c^3+\dots+x^{24}+y^{25}+z^{26}}{a^{26}+b^{25}+c^{24}+\dots + x^{3}+y^2+z} = \dfrac{\alpha}{\gamma}$

This works for all $$a,b,c,d,\dots , x,y,z$$ who either are not in decimal and not infinity. Take it as a challenge dude @Ark3 Graptor !

- 2 years, 11 months ago

no, its not a=1,b=2,c=3 etc. its this: a can equal 1,2,3,4,5....any other number, just not a=.1 or a=$\infty$

- 2 years, 11 months ago

Are you trying to argue the definition of a summation? This is not philosophy, I think you are in the wrong place.

- 2 years, 11 months ago

The series you are talking about does not converge for $$x\geq1$$.

- 2 years, 11 months ago

The variable number changes that so x=1, did you forget to read that part?

- 2 years, 11 months ago

If you don't mind, can you explain the concept of variable number to me with a few more examples?

- 2 years, 11 months ago

Ah, the variable number is designed for people to understand the true meaning and thought on a series, infinite or finite. This is what more than one variable numbers look like(the amount of sums can be infinite)

$\sum_{x=1}^\infty \sum_{y=2}^\infty \sum_{z=3}^\infty \frac{x+y+z}{z-y+x} = \frac{a}{b}=\frac{6}{2}$ The true meaning of a infinite series is to have no limit and for the number to not be changed

- 2 years, 11 months ago

If you don't mind can I talk to you on any other private platform(other than fb)? I need to ask you some questions.

- 2 years, 11 months ago

And converging isn't necessarily in this problem the only form of that is the dummy answer I edited in this morning

- 2 years, 11 months ago

What's the proof that your hypothesis is correct?

- 2 years, 11 months ago

@Ark3 Graptor This might help.

- 2 years, 11 months ago

Please read up on how to use the notation before posting. Using notation in a way that doesn't make sense might make you look smart among people who don't know what the notation means but it makes you look stupid among those who know.

- 2 years, 11 months ago

You are absolutely right.

- 2 years, 11 months ago