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!

## Comments

Sort by:

TopNewestI 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 \(5\). – Mark Hennings · 11 months, 1 week ago

Log in to reply

The correct formula for this series is the first formula

The rest made sense, thanks again – Ark3 Graptor · 11 months, 1 week ago

Log in to reply

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. – Mark Hennings · 11 months, 1 week ago

Log in to reply

– Ark3 Graptor · 11 months, 1 week 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 argumentLog in to reply

– Ark3 Graptor · 11 months, 1 week ago

Sorry,That was my bad excision of certain thingsLog in to reply

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. – Mark Hennings · 11 months, 1 week ago

Log in to reply

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

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 ! – Nihar Mahajan · 11 months, 1 week ago

Log in to reply

– Ark3 Graptor · 11 months, 1 week 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=\[\infty\]Log in to reply

– Julian Poon · 11 months, 1 week ago

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

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. – Julian Poon · 11 months, 1 week ago

Log in to reply

– Aditya Kumar · 11 months, 1 week ago

You are absolutely right.Log in to reply

@Ark3 Graptor This might help. – Ishan Singh · 11 months, 1 week ago

Log in to reply

What's the proof that your hypothesis is correct? – Aditya Kumar · 11 months, 2 weeks ago

Log in to reply

The series you are talking about does not converge for \(x\geq1\). – Aditya Kumar · 11 months, 2 weeks ago

Log in to reply

– Ark3 Graptor · 11 months, 1 week ago

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

– Aditya Kumar · 11 months, 1 week ago

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

\[\sum_{x=1}^\infty \sum_{y=2}^\infty \sum_{z=3}^\infty \frac{x+y+z}{z-y+x} = \frac{a}{b}=\frac{6}{2}\] The true meaning of a infinite series is to have no limit and for the number to not be changed – Ark3 Graptor · 11 months, 1 week ago

Log in to reply

– Aditya Kumar · 11 months, 1 week 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.Log in to reply

– Ark3 Graptor · 11 months, 1 week ago

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