n+1
nplusonemag.com
n+1 | n+1 is a print and digital magazine of literature, culture, and politics.
n+1 is a print and digital magazine of literature, culture, and politics published three times yearly. We post new online-only work several times each week and publish books expanding on the interests of the magazine.
About
n+1 is a print and digital magazine of literature, culture, and politics published three times yearly. We also post new online-only work several times each week and publish books expanding on the interests of the magazine.
Magazine
n+1 is a print and digital magazine of literature, culture, and politics published three times yearly.
Subscribe
“The best goddamn literary magazine in America.” —Mary Karr
Online Only
Get n+1 in your inbox.
Factsheet
Editors Lisa Borst, Mark Krotov, Dayna Tortorici
Categories Culture, Literature, Politics
Frequency Triannually
Editors Lisa Borst, Mark Krotov, Dayna Tortorici
Categories Culture, Literature, Politics
Frequency Triannually
Wikipedia
en.wikipedia.org › wiki › N+1
n+1 - Wikipedia
November 11, 2025 - n+1 is a New York–based American literary magazine that publishes social criticism, political commentary, essays, art, poetry, book reviews, and short fiction. It is published in print three times annually with regular articles being published online. Each print issue averages around 200 ...
A simple mental trick to quickly sum numbers from 1 to n
https://en.wikipedia.org/wiki/Triangular_number you can just do n(n+1)/2 More on reddit.com
13
0
October 30, 2025Why are factorials usually expressed as n! = n*(n-1)...*1 and not 1*2...*n?
Two reasons I can think of: It’s easier to represent factorial recursively as n! = n*(n-1)! = n* (n-1)* (n-2)* …. * 2 * 1. When you’re dealing with factorials in an expression, you often pull out the higher or first few highest terms and that’s easier to do when you represent factorials from largest to smallest. More on reddit.com
28
21
December 20, 2024How in the world does 1/n not converge?
Just to be clear, as n approaches infinity, 1/n converges to zero. What does not converge however, is infinite series of 1/n. There are several simple and intuitive proofs of this fact. For example, if you group the terms of the series as follows: 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16)...., It is clear that the sum of each group of terms in parentheses is greater than 1/2. We can then keep grouping the terms of the infinite series like this until we get a series which is greater than the infinite series of 1/2, and therefore must diverge. If you are confused why some series with nth terms equal to zero converge and others don't, you can use a simple tool from calculus is the p-series test. A p-series is a series of the form 1/np ,where p is a constant. The p-series test says that if p is less than or equal to 1, the infinite series of 1/np diverges, and if p is greater than 1, the series converges. For the infinite series of 1/n, p=1 and the series must diverge according to the p-series test. Hope this helps. More on reddit.com
28
125
July 4, 2020sequences and series - "Closed" form for $\sum \frac{1}{n^n}$ - Mathematics Stack Exchange
Earlier today, I was talking with my friend about some "cool" infinite series and the value they converge to like the Basel problem, Madhava-Leibniz formula for $\pi/4, \log 2$ and similar alternat... More on math.stackexchange.com
Videos
04:22
Real Analysis Course #15 - Limit of 1/n Converges to 0 - YouTube
06:48
Proof: Sequence 1/n Converges to 0 | Real Analysis Exercises - YouTube
06:43
Proof: Sequence (n+1)/n Converges to 1 | Real Analysis - YouTube
01:41
Root Test for Infinite Series SUM(1/n^n) - YouTube
08:13
Proof that 1^n = 1 for every integer n (First Proof) (ILIEKMAT...
How to Sum Up All Integers 1 to n using The Reverse Summation Method ...
Reddit
reddit.com › r/statistics › [q] proof question: where does the 1/n! come from?
r/statistics on Reddit: [Q] Proof question: where does the 1/n! come from?
August 18, 2022 -
So I’m working out the problem “sum greater than 1” where you have to find the tipping point where an amount of random numbers (0,1) sum to greater than one.
In my journey to asking ‘why’, I’m curious, where does the 1/n! come from? It seems like it sort of pops out of no where and am curious how you might be able to prove how 1/n! works through integration/derivation.
I think it has something to do with the law of iterated expectations, but feel like I’m missing something.
Here are a few formulas I’m using as reference
https://imgur.com/gallery/TI5taYI
Top answer 1 of 4
6
X=1 in the top formula
2 of 4
3
I’m not sure if I’m answering what you’re looking for here, as it doesn’t really relate to the “sum greater than 1” thing directly, but are you asking where the 1/n! comes from in the Taylor expansion? The 3blue1brown video I linked below explains it better than I can, but the general idea is you’re finding the Taylor expansion of ex centered around zero and plugging in 1. For Taylor series in general, you’re essentially trying to fit a function to an infinite polynomial. You do this by fitting your polynomial to all the higher order derivatives of your function, and because of the way the power rule works for the polynomial, you end up with factorials. Not gonna type out details, but the video should help if that’s the part you’re confused about. https://youtu.be/3d6DsjIBzJ4
N+1 Bikes
n1bikes.com
N+1 Bikes | Louisville, KY
Looking for something unique? N+1 Bikes specializes in custom builds, tailoring your bike to perfection and shipping it fully assembled right to your front door.
Wikipedia
en.wikipedia.org › wiki › Ramsey's_theorem
Ramsey's theorem - Wikipedia
December 20, 2025 - More precisely, the theorem states that for any given number of colours, c, and any given integers n1, …, nc, there is a number, R(n1, …, nc), such that if the edges of a complete graph of order R(n1, …, nc) are coloured with c different colours, then for some i between 1 and c, it must contain a complete subgraph of order ni whose edges are all colour i.
Boston University
bu.edu › housing › wp-content › themes › r-housing › js › vendor › pannellum › pannellum.htm
N=1 But Here's My Data on Balance Your Blood Sugar to ...
Javascript is required to view this panorama. (It could be worse; you could need a plugin.)
Facebook
facebook.com › nplusonemag
n+1 | Facebook
n+1. 75,394 likes · 3 talking about this. n+1 is a print and digital magazine of literature, culture, and politics, based in New York. See our website at nplusonemag.com.
Reddit
reddit.com › r/askmath › a simple mental trick to quickly sum numbers from 1 to n
r/askmath on Reddit: A simple mental trick to quickly sum numbers from 1 to n
October 30, 2025 -
I want to share a simple and visual formula for summing numbers from 1 to n, which I invented. It allows you to see the pattern and quickly calculate sums without a calculator.
Top answer 1 of 5
11
https://en.wikipedia.org/wiki/Triangular_number you can just do n(n+1)/2
2 of 5
3
visual How is that visual, exactly? There's no visual support. simple I guess it is simple in the grand scheme of things, but calling it simple is disingenuous when there is an equivalent formula that's way simpler, i.e. (n+1)n/2. When n is even, ceil(n/2)=n/2 and your formula reduces to (n+1)n/2. When n is odd, ceil(n/2)=(n+1)/2 and the formula still reduces to (n+1)n/2. All your method does is obfuscate this simple formula and add awful notation (who writes their cases in words in a parenthesis along with the factor?). which I invented Not so. This method is often credited to Carl Friedrich Gauss in the late 18th century. Edit: I conflated two stories. Carl Friedrich Gauss might've figured it out in the late 18th century, but it was invented way earlier than that. I should've realized before posting, as there's no way it's so recent.
Chill Subs
chillsubs.com › magazine › nplusonemag
n+1
No AI. No ads. Mostly free. Join 69,000+ registered writers and discover 4235 publishing opportunities.
Wikipedia
en.wikipedia.org › wiki › Newton_(unit)
Newton (unit) - Wikipedia
1 month ago - The newton (symbol: N) is the unit of force in the International System of Units (SI). Expressed in terms of SI base units, it is 1 kg⋅m/s2, the force that accelerates a mass of one kilogram at one metre per second squared.
Reddit
reddit.com › r/askmath › why are factorials usually expressed as n! = n*(n-1)...*1 and not 1*2...*n?
r/askmath on Reddit: Why are factorials usually expressed as n! = n*(n-1)...*1 and not 1*2...*n?
December 20, 2024 -
Is it just tradition or is there an actual reason?
Top answer 1 of 11
69
Two reasons I can think of: It’s easier to represent factorial recursively as n! = n*(n-1)! = n* (n-1)* (n-2)* …. * 2 * 1. When you’re dealing with factorials in an expression, you often pull out the higher or first few highest terms and that’s easier to do when you represent factorials from largest to smallest.
2 of 11
26
Because usually, when manipulating factorials, you'll unfold "n" out of it so it's more natural to think of "n" as the first factor.
Reddit
reddit.com › r/learnmath › how in the world does 1/n not converge?
r/learnmath on Reddit: How in the world does 1/n not converge?
July 4, 2020 -
I like to come to things with an intuitive approach, but with 1/n I just can't come to terms with it! My understanding of convergence is that you add an infinite amount of values for the function and it gets closer and closer to a finite value. With 1/n as n gets towards infinity the output becomes infinitely small also, so why is it not approaching a value?
Top answer 1 of 14
141
Just to be clear, as n approaches infinity, 1/n converges to zero. What does not converge however, is infinite series of 1/n. There are several simple and intuitive proofs of this fact. For example, if you group the terms of the series as follows: 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16)...., It is clear that the sum of each group of terms in parentheses is greater than 1/2. We can then keep grouping the terms of the infinite series like this until we get a series which is greater than the infinite series of 1/2, and therefore must diverge. If you are confused why some series with nth terms equal to zero converge and others don't, you can use a simple tool from calculus is the p-series test. A p-series is a series of the form 1/np ,where p is a constant. The p-series test says that if p is less than or equal to 1, the infinite series of 1/np diverges, and if p is greater than 1, the series converges. For the infinite series of 1/n, p=1 and the series must diverge according to the p-series test. Hope this helps.
2 of 14
82
One way to see it fairly intuitively is to note that 1/1 1/2 1/3 + 1/4 1/5 + 1/6 + 1/7 + 1/8 and so on are each equal to at least 1/2. Therefore, whatever partial sum you reach, you can always increase the sum by at least 1/2 by simply using twice as many terms. And since you can just keep doubling the number of terms in your partial sum, you can keep increasing the partial sum's value by 1/2, ad infinitum. Edit: typos.
Top answer 1 of 4
57
Certainly you need to check out Sophomore's Dream.
2 of 4
29
This series turns out to be equal to the integral of $1/x^x$ from $0$ to $1$. Sorry I am bad with the notation needed for math symbols here; but if you rewrite $x^{-x}$ as the exponential of $x\log x$, you can expand it as a taylor series and integrate term by term; to find it equals your sum. This doesn't give an explicit value; but I think it's a pretty cool identity.
EDIT: Just putting what you said into symbols, probably for the fun of it.
$$\int_0^1 x^{-x} dx=\int_0^1 e^{-x \log x} dx$$
$$ e^{-x \log x} =\sum_{k=0}^\infty (-1)^k \frac{x^k\log^kx}{k!}$$
But since
$$\int_0^1 x^k \log^k x dx =(-1)^k \frac{k!}{(k+1)^{k+1}}$$
we get
$$\int_0^1 x^{-x} dx= \sum_{k=0}^\infty \frac{1}{(k+1)^{k+1}}=\sum_{k=1}^\infty \frac{1}{k^k}$$
Goodreads
goodreads.com › book › show › 70578.n_1_Issue_1
n+1 Issue 1: Negation by n+1 | Goodreads
n+1 is a print journal of politics, literature, and culture, published originally twice a year and now three times a year.
Authors n+1Sam Lipsyte…
Pages 182
N1labs
n1labs.org
The N.1 Institute for Health
N.1 is home to a leading team of experts in neuroengineering, harnessing the role of training, machine learning, and connectivity imaging to predict and ultimately augment human cognitive performance.
Python
docs.python.org › 3 › library › itertools.html
itertools — Functions creating iterators for efficient looping
1 week ago - def count(start=0, step=1): # count(10) → 10 11 12 13 14 ... # count(2.5, 0.5) → 2.5 3.0 3.5 ... n = start while True: yield n n += step
LinkedIn
linkedin.com › company › nplusonefoundation
n+1 Foundation | LinkedIn
n+1 Foundation | 897 followers on LinkedIn. Nonprofit publisher of n+1 magazine, n+1 Books, and Paper Monument. | More than a decade on, n+1 has “established itself as the bellwether of a new generation of literary intellectuals” (Harper’s). It has published dozens of book-length magazines, hundreds of online-only pieces, and a handful of critically acclaimed books, as well as programmed countless readings, panels, and events across New York City.
Convert Me
convert-me.com › measurement conversion › engineering converters › force units converter › international system (si) › newton
N - Newton. Conversion Chart / Force Units Converter, International System (SI)
Force Units Converter / International System (SI) / Newton [N] Online converter page for a specific unit. Here you can make instant conversion from this unit to all other compatible units.
Wikipedia
en.wikipedia.org › wiki › 1
1 - Wikipedia
3 weeks ago - It is the first and smallest positive integer of the infinite sequence of natural numbers. This fundamental property has led to its unique uses in other fields, ranging from science to sports, where it commonly denotes the first, leading, ...
Uwyo
journals.uwyo.edu › index.php › ela › article › view › 8961 › 7373
View of Accurate computation of all eigenvalues of a totally nonnegative matrix by the Cauchon algorithm
Return to Article Details Accurate computation of all eigenvalues of a totally nonnegative matrix by the Cauchon algorithm Download Download PDF