CristianTheodor
New member
- Joined
- Jul 14, 2019
- Messages
- 22
I cannot really read the attachment. But look at this.
Yes , I'm wrong . But how to solve it ?I cannot really read the attachment. But look at this.
I don't think I actually follow any of the work you've shown. You started by taking the summation expression and multiplying it by 2?? I don't understand the purpose of this step. Further, you state that:
3nn+1⋅2=3n+1n+2
But this is only true if n=−54, so that doesn't seem to help you any. You then somehow use this to derive a mysterious expression for an that is completely incorrect and seems to appear from nowhere. The only thing this working shows correctly is taking the limit - If that were a valid expression for an, then the limit would be 2.
If nothing else, we can see that the limit must be incorrect, as the an is a strictly increasing sequence (why?) and the terms already exceed 2 when n=3:
a3=k=0∑33kk+1=300+1+311+1+322+1+333+1=2758>2
To be honest, your bizarre and seemingly disjointed work makes me wonder if you're familiar with sigma notation. Is that the case? If it is, I've linked a page where you can refresh yourself on this notation.
That's not 2 , it's q . I use the geometric series sum . The q is the ratio .I don't think I actually follow any of the work you've shown. You started by taking the summation expression and multiplying it by 2?? I don't understand the purpose of this step. Further, you state that:
3nn+1⋅2=3n+1n+2
But this is only true if n=−54, so that doesn't seem to help you any. You then somehow use this to derive a mysterious expression for an that is completely incorrect and seems to appear from nowhere. The only thing this working shows correctly is taking the limit - If that were a valid expression for an, then the limit would be 2.
If nothing else, we can see that the limit must be incorrect, as the an is a strictly increasing sequence (why?) and the terms already exceed 2 when n=3:
a3=k=0∑33kk+1=300+1+311+1+322+1+333+1=2758>2
To be honest, your bizarre and seemingly disjointed work makes me wonder if you're familiar with sigma notation. Is that the case? If it is, I've linked a page where you can refresh yourself on this notation.
If ∣x∣<1 then this is a well known sum S(x)=k=0∑∞xk=1−x1I don't think I actually follow any of the work you've shown. You started by taking the summation expression and multiplying it by 2?? I don't understand the purpose of this step. Further, you state that:
That's so clever.Thank you so much .If ∣x∣<1 then this is a well known sum S(x)=k=0∑∞xk=1−x1
We multiply by x in order to get k+1 into the sum x⋅S(x)=k=0∑∞xk+1=1−xx
Finding the derivative of that we get
S(x)+x⋅S′(x)=k=0∑∞(k+1)xk=(1−x)21
Let x=31 or k=0∑∞3kk+1=(1−31)21=49
Being right or wrong is NOT the real problem. You and PKA seem to be solving different problems! PKA is solving a summation (which I think that you want) WHILE you are solving a limit. Why is this?Yes , I'm wrong . But how to solve it ?
To Jomo, are you really serious?Being right or wrong is NOT the real problem. You and PKA seem to be solving different problems! PKA is solving a summation (which I think that you want) WHILE you are solving a limit. Why is this?
Read my bold comments above.To Jomo, are you really serious?
Surely you know what it what it means for a infinite series to converse means? Don't you??An infinite series converges if the infinite sum is finite
Suppose that an=k=0∑nf(k) then the sequence an is a numerical sequence of partial sums. I understand that.
If the sequence converges an→S then we say k=0∑∞f(k)=S. Sure
Jomo, did you read the original postings? In it CristianTheodor asked about the partial sum an=k=1∑n3kk+1 Yes he did
Do you understand summations of infinite series? If not please don't say that I miss-read the question.
Look at reply #9 and then tell me that I am mistaken. Never said you were mistaken